Arithmetic Ability Mock Test

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This Mock Test is created questions of Arithmetic Ability, also known as Quantitative Aptitude or Numerical Ability from Previous Year Recruitment Papers by our experts.

Answered
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Question 1 of 50

1. Question
1 points
The respective ratio between the present ages of Ram,Rohan and Raj is 3 : 4 : 5. if the average of their presentages is 28 yr then what would be the sum of the ages of
Ram and Rohan together after 5 yr?
- 1. 45 yr
- 2. 55 yr
- 3. 52 yr
- 4. 59 yr
- 5. None of these
Correct

Let the ages of Ram, Rohan and Raj be 3 x,4 x and 5 x
respectively,
Then,

$\frac{3 x+4 x+5 x}{3}$ =28

$\Rightarrow$ 4 x=28

$\Rightarrow x=\frac{28}{4}$ =7 yr

So, the present ages of Ram and Rohan together

=3 x+4 x

=7 x= $7\times 7$

= 49 yr

Hence, the sum of ages of Ram and Rohan together
after 5 yr

=49+ $5\times 7$

=49+10

=59 yr

Incorrect

Let the ages of Ram, Rohan and Raj be 3 x,4 x and 5 x
respectively,
Then,

$\frac{3 x+4 x+5 x}{3}$ =28

$\Rightarrow$ 4 x=28

$\Rightarrow x=\frac{28}{4}$ =7 yr

So, the present ages of Ram and Rohan together

=3 x+4 x

=7 x= $7\times 7$

= 49 yr

Hence, the sum of ages of Ram and Rohan together
after 5 yr

=49+ $5\times 7$

=49+10

=59 yr
Question 2 of 50

2. Question
1 points
Raman scored 456 marks in an exam and Seeta got54 per cent marks in the same exam which is 24 marks lessthan Raman. If the minimum passing marks in the exam is34 percent,then how much more marks did Raman score than the minimum passing marks?
- 1. 184
- 2. 196
- 3. 190
- 4. 180
- 5. None of these
Correct

Let the total marks of the exam be x.
Then,

$x\times \frac{54}{100}$ =456-24

$\Rightarrow �x\times \frac{54}{100}$ =432

$\Rightarrow �x\times \frac{432\times 100}{54}$ =800

$\therefore$ Minimum passing marks

= $�800\times \frac{34}{100}$

=272

Hence, required more marks get by Raman
= 456 -272
= 184

Incorrect

Let the total marks of the exam be x.
Then,

$x\times \frac{54}{100}$ =456-24

$\Rightarrow �x\times \frac{54}{100}$ =432

$\Rightarrow �x\times \frac{432\times 100}{54}$ =800

$\therefore$ Minimum passing marks

= $�800\times \frac{34}{100}$

=272

Hence, required more marks get by Raman
= 456 -272
= 184
Question 3 of 50

3. Question
1 points
In the following questions: two equations numbered I ond II are given. You have to :. solve both the equations and-

I. $(x^{\frac{7}{5}}\div 9)=169\div x^{\frac{3}{5}}$

II. $y^{\frac{1}{4}}\times y^{\frac{1}{4}}\times 7=273\div y^{\frac{1}{2}}$
- 1. if x > y
- 2. if x ≥ y
- 3. if x < y
- 4. if x ≤ y
- 5. if x=y or the relationship cannot be established
Correct

From Eq .(i),

$\left ( x^{\frac{7}{5}}\div 9 \right )=169\div x^{\frac{3}{5}}$

$\Rightarrow x^{\frac{7}{5}+^{\frac{3}{5}}}=9\times 169$

$\Rightarrow x^{2}=1521$

$\Rightarrow x=\sqrt{1521}=\pm 39$

From Eq. (ii),

$y^{\frac{1}{4}}\times y^{\frac{1}{4}}\times 7=273\div y^{\frac{1}{2}}$

$\Rightarrow y(^{\frac{1}{4}}+^{\frac{1}{4}})-\frac{1}{2}=\frac{273}{7}$

$\Rightarrow y^{0}=39$

$\Rightarrow$ y=39

Hence, x

≤ y

Incorrect

From Eq .(i),

$\left ( x^{\frac{7}{5}}\div 9 \right )=169\div x^{\frac{3}{5}}$

$\Rightarrow x^{\frac{7}{5}+^{\frac{3}{5}}}=9\times 169$

$\Rightarrow x^{2}=1521$

$\Rightarrow x=\sqrt{1521}=\pm 39$

From Eq. (ii),

$y^{\frac{1}{4}}\times y^{\frac{1}{4}}\times 7=273\div y^{\frac{1}{2}}$

$\Rightarrow y(^{\frac{1}{4}}+^{\frac{1}{4}})-\frac{1}{2}=\frac{273}{7}$

$\Rightarrow y^{0}=39$

$\Rightarrow$ y=39

Hence, x

≤ y
Question 4 of 50

4. Question
1 points
In the following questions: two equations numbered I and II are given. You have to :. solve both the equations and-

II. $4y^{3}=(589\div 4)+5y^{3}$
- 1. if x >y
- 2. if x ≥ y
- 3. if x < y
- 4. if x ≤ y
- 5. if x = y or the relationship cannot be established
Correct

From Eq. (i),

$x^{3}=\frac{(2)^{5}+(11)^{3}}{6}$

$\Rightarrow x^{3}=\frac{32+1331}{6}$

$\Rightarrow x^{3}=\frac{1363}{6}$ = 227.17

$\Rightarrow x=3\sqrt{227.17}$

From Eq. (ii),

$4y^{3}=(-589\div 4)+54^{3}$

$\Rightarrow y^{3}=\frac{589}{4}$ =147.25

$\Rightarrow y=3\sqrt{147.25}$

Hence, x > y

Incorrect

From Eq. (i),

$x^{3}=\frac{(2)^{5}+(11)^{3}}{6}$

$\Rightarrow x^{3}=\frac{32+1331}{6}$

$\Rightarrow x^{3}=\frac{1363}{6}$ = 227.17

$\Rightarrow x=3\sqrt{227.17}$

From Eq. (ii),

$4y^{3}=(-589\div 4)+54^{3}$

$\Rightarrow y^{3}=\frac{589}{4}$ =147.25

$\Rightarrow y=3\sqrt{147.25}$

Hence, x > y
Question 5 of 50

5. Question
1 points
The cost of five chairs and three tables is Rs.3110.Cost of one chair is Rs. 210 less than cost of one table.What is cost of two tables and two chairs?
- 1. Rs.1660
- 2. Rs.1860
- 3. Rs.2600
- 4. Cannot be determined
- 5. None of these
Correct

Suppose the cost of one Chair = C
and the cost of one Table = ?
Then,
5 C+3'T=3110 ....(i)
and
T -C = 210 or T =210 + c ...(ii)
On putting value of T in Eq. (i),
5 c+3(210+c)=3110

$\Rightarrow$ 5 C+630+3 C=3110

$\Rightarrow$ 8 C =3110-630

$\Rightarrow C= \frac{2480}{8}$ =Rs.310

$\therefore$ Cost of one Table (T) = 210 + 310
= Rs. 520

Hence, the cost of two Tables and two Chairs.
=2 T +2 C
= $2\times 520+2\times 310$

= 1040 + 620
=Rs.1660

Incorrect

Suppose the cost of one Chair = C
and the cost of one Table = ?
Then,
5 C+3'T=3110 ....(i)
and
T -C = 210 or T =210 + c ...(ii)
On putting value of T in Eq. (i),
5 c+3(210+c)=3110

$\Rightarrow$ 5 C+630+3 C=3110

$\Rightarrow$ 8 C =3110-630

$\Rightarrow C= \frac{2480}{8}$ =Rs.310

$\therefore$ Cost of one Table (T) = 210 + 310
= Rs. 520

Hence, the cost of two Tables and two Chairs.
=2 T +2 C
= $2\times 520+2\times 310$

= 1040 + 620
=Rs.1660
Question 6 of 50

6. Question
1 points
$\frac{1}{6}of(92)$ %of 1 $\frac{1}{23}$ of(650)=85+?
- 1. 18
- 2. 21
- 3. 19
- 4. 28
Correct

$85+?=\frac{1}{6}\times\frac{92}{100}\times \frac{24}{23}\times 650$
85+?=104
?=104-85
?=19

Incorrect

$85+?=\frac{1}{6}\times\frac{92}{100}\times \frac{24}{23}\times 650$
85+?=104
?=104-85
?=19
Question 7 of 50

7. Question
1 points
92 x 576÷2 $\sqrt{1296}$ =(?)³ + √49
- 1. 3
- 2. (9)²
- 3. 9
- 4. 27
- 5. None of these
Correct

(?)³+7=92×576÷2×36

(?)³+7=92×576÷72

(?)³+7=92×576÷72

(?)³+7=92×8

(?)³+7=736

(?)³=736-7

?=³√729

?=9

Incorrect

(?)³+7=92×576÷2×36

(?)³+7=92×576÷72

(?)³+7=92×576÷72

(?)³+7=92×8

(?)³+7=736

(?)³=736-7

?=³√729

?=9
Question 8 of 50

8. Question
1 points
$3\frac{1}{4}+2\frac{1}{2}-1\frac{5}{6}=\frac{? ^{2}}{10} +1\frac{5}{12}$
- 1. 25
- 2. √5
- 3. 625
- 4. 15
- 5. 5
Correct

Incorrect
Question 9 of 50

9. Question
1 points
$\left ( \sqrt{8}\times \sqrt{8} \right )^{\frac{1}{2}}+\left ( 9\right )^{^{\frac{1}{2}}}=\sqrt{8}-340$
- 1. 7
- 2. 19
- 3. 18
- 4. 9
- 5. None of these
Correct

$(1)\left ( ?\right )^{3}+\left \sqrt{8}-340=(\sqrt{8}\times \sqrt{8})^{\frac{1}{2}}+\left ( 9 \right )^{\frac{1}{2}}$

$\left ( ? \right )^{3}+\sqrt{8}-340=\sqrt{8}+3$

$\left ( ? \right )^{3}=\sqrt{8}+3-\sqrt{8}+340$

$\left ( ? \right )=\sqrt[3]{343}$

?=7

Incorrect

$(1)\left ( ?\right )^{3}+\left \sqrt{8}-340=(\sqrt{8}\times \sqrt{8})^{\frac{1}{2}}+\left ( 9 \right )^{\frac{1}{2}}$

$\left ( ? \right )^{3}+\sqrt{8}-340=\sqrt{8}+3$

$\left ( ? \right )^{3}=\sqrt{8}+3-\sqrt{8}+340$

$\left ( ? \right )=\sqrt[3]{343}$

?=7
Question 10 of 50

10. Question
1 points
$\left ( \frac{24}{9} \right )^{2}\times \frac{399}{39}\div \frac{41}{899}=?$
- 1. 1600
- 2. 1650
- 3. 1700
- 4. 1550
- 5. 1750
Correct

$\left ( \frac{24}{9} \right )^{2}\times \frac{399}{39}\div \frac{41}{899}$

$?=\left ( \frac{24}{9} \right )^{2}\times \frac{399}{39}\times \frac{899}{41}$

$? =7.11\times 10.23\times 21.92$

?=1594.35 = 1550 Approx.

Incorrect

$\left ( \frac{24}{9} \right )^{2}\times \frac{399}{39}\div \frac{41}{899}$

$?=\left ( \frac{24}{9} \right )^{2}\times \frac{399}{39}\times \frac{899}{41}$

$? =7.11\times 10.23\times 21.92$

?=1594.35 = 1550 Approx.
Question 11 of 50

11. Question
1 points
$(15\times 0.40)^{4}\div (1080\div 30)^{4}\times (27\times 8)^{4}=(3\times 2)^{?+5}$
- 1. 8
- 2. 3
- 3. 12
- 4. 16
- 5. None of these
Correct

$\left ( 3\times 2 \right )^{?+5} = \left ( 15\times 0.40\right )^{4}\div \left ( 1080\div 30 \right )^{4}\times \left ( 27\times 8\right )^{4}$

$\left ( 3\times 2 \right )^{?+5}=\left ( 6 \right )^{4}\div \left ( 36 \right )^{4}\times \left ( 216 \right )^{4}$

$\left ( 6 \right )^{?+5}$ = $(6)^{4}\div (6^{2})^{4}\times (6^{3})^{4}$

$\left ( 6 \right )^{?+5}=\left ( 6 \right )^{4}\times \left ( 6 \right )^{8}\times \left ( 6 \right )^{12}$

$\left ( 6 \right )^{?+5}=\left ( 6 \right )^{-4}\times \left ( 6 \right )^{12}$

$\left ( 6 \right )^{?+5}=\left ( 6 \right )^{8}$

? + 5 = 8
? = 8 - 5 = 3

Incorrect

$\left ( 3\times 2 \right )^{?+5} = \left ( 15\times 0.40\right )^{4}\div \left ( 1080\div 30 \right )^{4}\times \left ( 27\times 8\right )^{4}$

$\left ( 3\times 2 \right )^{?+5}=\left ( 6 \right )^{4}\div \left ( 36 \right )^{4}\times \left ( 216 \right )^{4}$

$\left ( 6 \right )^{?+5}$ = $(6)^{4}\div (6^{2})^{4}\times (6^{3})^{4}$

$\left ( 6 \right )^{?+5}=\left ( 6 \right )^{4}\times \left ( 6 \right )^{8}\times \left ( 6 \right )^{12}$

$\left ( 6 \right )^{?+5}=\left ( 6 \right )^{-4}\times \left ( 6 \right )^{12}$

$\left ( 6 \right )^{?+5}=\left ( 6 \right )^{8}$

? + 5 = 8
? = 8 - 5 = 3
Question 12 of 50

12. Question
1 points
67.99% of 1401 - 13.99% of 1299 = ?
- 1. 700
- 2. 720
- 3. 770
- 4. 800
- 5. 740
Correct

? = 6799% of 1401 -13.99 of 1299
? = $1401\times \frac{68}{100}-1300\times \frac{14}{100}$
? = 952.68 - 182
? = 770.68
? = 770 (Approx.)

Incorrect

? = 6799% of 1401 -13.99 of 1299
? = $1401\times \frac{68}{100}-1300\times \frac{14}{100}$
? = 952.68 - 182
? = 770.68
? = 770 (Approx.)
Question 13 of 50

13. Question
1 points
5466.97 - 3245.01 + 1122.99 = ?+ 2309.99
- 1. 1130
- 2. 10000
- 3. 1100
- 4. 1030
- 5. 1060
Correct

? = 2309.99 = 5466.97 - 3245.01+ 1122.99
? + 2310 = 5467 - 3245 + 1123
? = 5467 - 3245 + 1123 - 2310
? = 5467 +1123 - (3245+2310)
? = 6590 - 5555
? = 1035
? = 1030 (Approx.)

Incorrect

? = 2309.99 = 5466.97 - 3245.01+ 1122.99
? + 2310 = 5467 - 3245 + 1123
? = 5467 - 3245 + 1123 - 2310
? = 5467 +1123 - (3245+2310)
? = 6590 - 5555
? = 1035
? = 1030 (Approx.)
Question 14 of 50

14. Question
1 points
5998 $\div$ 9.98 + 670.99 - 139.99 = ?
- 1. 1080
- 2. 1280
- 3. 1180
- 4. 1130
- 5. 1230
Correct

? = 5998 + 9.98 + 670.99 - 139.99
? = 5998 $\div$ 10 + 671 - 140
? = 599.8 + 531
? = 1130.8
? =1130 (Approx.)

Incorrect

? = 5998 + 9.98 + 670.99 - 139.99
? = 5998 $\div$ 10 + 671 - 140
? = 599.8 + 531
? = 1130.8
? =1130 (Approx.)
Question 15 of 50

15. Question
1 points
$- (4.99)^{3} +(29.98)^{2}-(3.01)^{4}=?$
- 1. 550
- 2. 590
- 3. 620
- 4. 650
- 5. 690
Correct

$(5)^{?}=(499)^{3}+(29.98)^{2}-(3.01)^{4}$
$?= -(5)^{3}+(30)^{2}-(3)^{4}$
? = - 125 + 900 - 81
? = 900 - 206
? = 694
? = 690 (Approx.)

Incorrect

$(5)^{?}=(499)^{3}+(29.98)^{2}-(3.01)^{4}$
$?= -(5)^{3}+(30)^{2}-(3)^{4}$
? = - 125 + 900 - 81
? = 900 - 206
? = 694
? = 690 (Approx.)
Question 16 of 50

16. Question
1 points
1 7 49 343 (?)
- 1. 16807
- 2. 1227
- 3. 2058
- 4. 2401
- 5. None of these
Correct

The sequence of the series is:

Incorrect

The sequence of the series is:
Question 17 of 50

17. Question
1 points
What will come in place of question mark (?) in the following number series?

1 3 20 39 78 145 (?)
- 1. 234
- 2. 244
- 3. 236
- 4. 248
- 5. None of these
Correct

Incorrect
Question 18 of 50

18. Question
1 points
What will come in place of question mark (?) in the following number series ?

12 35 81 173 357 (?)
- 1. 725
- 2. 715
- 3. 726
- 4. 736
- 5. None of these
Correct

Incorrect
Question 19 of 50

19. Question
1 points
What will come in place of question mark (?) in the following number series?

3 100 297 594 991 (?)
- 1. 1489
- 2. 1479
- 3. 1478
- 4. 1498
- 5. None of these
Correct

3+97=100
+100 $\downarrow$
100 + 197 =297
+100 $\downarrow$
297 + 297 =594
+100 $\downarrow$
594+397=991
+100 $\downarrow$
991+497=1488

Incorrect

3+97=100
+100 $\downarrow$
100 + 197 =297
+100 $\downarrow$
297 + 297 =594
+100 $\downarrow$
594+397=991
+100 $\downarrow$
991+497=1488
Question 20 of 50

20. Question
1 points
What will come in place of question mark (?) in the following number series?

112 119 140 175 224 (?)
- 1. 277
- 2. 276
- 3. 287
- 4. 266
- 5. None of these
Correct

Incorrect
Question 21 of 50

21. Question
1 points
In the following questions two equations numbered I and II are given. You have to :. solve both the equations and

I. $\sqrt{1225}x+\sqrt{4900}$
II. $(81)^{\frac{1}{4}}y+(343)^{\frac{1}{3}}$
- 1. if x > y
- 2. if x ≥ y
- 3. if x < y
- 4. if x ≤ y
- 5. if x = y or the relationship cannot be established
Correct

From Eq. (i), $\sqrt{1225}x+\sqrt{4900}$ =0
$\Rightarrow$ 35 x+70=0
$\Rightarrow$ x=-2
From Eq. (ii),
$81\tfrac{1}{4}y+343\tfrac{1}{3}$ =0
$\Rightarrow$ 3y+7=0
$y=\frac{-7}{3}=-2\frac{1}{3}$
Hence,X=Y

Incorrect

From Eq. (i), $\sqrt{1225}x+\sqrt{4900}$ =0
$\Rightarrow$ 35 x+70=0
$\Rightarrow$ x=-2
From Eq. (ii),
$81\tfrac{1}{4}y+343\tfrac{1}{3}$ =0
$\Rightarrow$ 3y+7=0
$y=\frac{-7}{3}=-2\frac{1}{3}$
Hence,X=Y
Question 22 of 50

22. Question
1 points
In the following questions: two equations numbered I and II are given. You have to :. solve both the equations and-

I. $\frac{18}{^{x2}}+\frac{6}{x}-\frac{12}{^{x2}}=\frac{8}{_{x2}}$
II. $^{y3}$ +9.68+5.64=16.95
- 1. if x > y
- 2. if x ≥ y
- 3. if x < y
- 4. if x ≤ y
- 5. if x=y or the relationship cannot be established
Correct

From Eq. (i),
$\frac{18}{^{x2}}+\frac{6}{^{x2}}-\frac{12}{^{x2}}=\frac{8}{^{x2}}$
$\Rightarrow \frac{6}{^{x2}}+\frac{6}{x}-\frac{8}{^{x2}}$ =0
$\Rightarrow -\frac{2}{^{x2}}+\frac{6}{x}$ =0
$\Rightarrow \frac{2}{^{x2}}=\frac{6}{x}$
$\Rightarrow x=\frac{2}{6}=\frac{1}{3}$ =0.33
From Eq. (ii),
$^{y3}$ +9.68+5.64=16.95
$\Rightarrow ^{y3}$ =16.95-(9.68+5.64)
$\Rightarrow ^{y3}$ =16.95-15.32
$\Rightarrow ^{y3}$ =1.63
$\Rightarrow y= 3\sqrt{1.63}$ = 1.176
Hence,y > x

Incorrect

From Eq. (i),
$\frac{18}{^{x2}}+\frac{6}{^{x2}}-\frac{12}{^{x2}}=\frac{8}{^{x2}}$
$\Rightarrow \frac{6}{^{x2}}+\frac{6}{x}-\frac{8}{^{x2}}$ =0
$\Rightarrow -\frac{2}{^{x2}}+\frac{6}{x}$ =0
$\Rightarrow \frac{2}{^{x2}}=\frac{6}{x}$
$\Rightarrow x=\frac{2}{6}=\frac{1}{3}$ =0.33
From Eq. (ii),
$^{y3}$ +9.68+5.64=16.95
$\Rightarrow ^{y3}$ =16.95-(9.68+5.64)
$\Rightarrow ^{y3}$ =16.95-15.32
$\Rightarrow ^{y3}$ =1.63
$\Rightarrow y= 3\sqrt{1.63}$ = 1.176
Hence,y > x
Question 23 of 50

23. Question
1 points
In the following questions: two equations numbered I and II are given. You have to :. solve both the equations and-

I. $12^{x2}+11x+12=10x^{2}+22x$
II. $13y^{3}-18y+3=9y^{2}-10y$
- 1. if x > y
- 2. if x ≥ y
- 3. if x < y
- 4. if x ≤ y
- 5. if x =y or the relationship cannot be established
Correct

From Eq. (i),
$12^{x2}+11x+12=10^{x2}+22x$
$\Rightarrow 2x^{2} -11x+12=0$
$\Rightarrow 2x^{2} -8x-3x +12=0$
$\Rightarrow 2x(x-4)-3(x-4)=0$
$\Rightarrow (2x-3)(x-4)=0$
Then,
(2x-3)=0 $\Rightarrow x=\frac{3}{2}$
and
(x-4)=0 \Rightarrow x = 4
From Eq. (ii),
$13y^{2}-18y+3=9y^{_{3}}-10y$
$\Rightarrow 4y-8y+3=0$
$\Rightarrow 4y-6y-2y+3=0$
$\Rightarrow 2y(2y-3)-1(2y-3)= 0$
$\Rightarrow (2y-1)(2y-3)= 0$
Then,
(2y-1)=0 $\Rightarrow y=\frac{1}{2}$
and
(2y-3)=0 $\Rightarrow y=\frac{3}{2}$
Hence, x ≥ y

Incorrect

From Eq. (i),
$12^{x2}+11x+12=10^{x2}+22x$
$\Rightarrow 2x^{2} -11x+12=0$
$\Rightarrow 2x^{2} -8x-3x +12=0$
$\Rightarrow 2x(x-4)-3(x-4)=0$
$\Rightarrow (2x-3)(x-4)=0$
Then,
(2x-3)=0 $\Rightarrow x=\frac{3}{2}$
and
(x-4)=0 \Rightarrow x = 4
From Eq. (ii),
$13y^{2}-18y+3=9y^{_{3}}-10y$
$\Rightarrow 4y-8y+3=0$
$\Rightarrow 4y-6y-2y+3=0$
$\Rightarrow 2y(2y-3)-1(2y-3)= 0$
$\Rightarrow (2y-1)(2y-3)= 0$
Then,
(2y-1)=0 $\Rightarrow y=\frac{1}{2}$
and
(2y-3)=0 $\Rightarrow y=\frac{3}{2}$
Hence, x ≥ y
Question 24 of 50

24. Question
1 points
The total area of a circle and a rectangle is equal to 1166 sq cm. The diameter of the circle is 28 cm. What is the sum of the circumference of the circle and the perimeter of the
rectangle if the length of the rectangle is 25 cm?
- 1. 186 cm
- 2. 182 cm
- 3. 184 cm
- 4. Cannot be determined
- 5. None of these
Correct

The area of the circle= $\pi r^{2}$

= $\frac{22}{7}\times 14\times 14$

=616 sq cm

Let the length and breadth of the rectangle be x cm
and y cm respectively.
Then,

$x\times y$ =1166-616

$\Rightarrow 25\times y$ =550

$\Rightarrow y=\frac{550}{25}$ =22 cm

$\therefore$ Perimeter of the rectangle

=2(x+y)

=2(25+22)

=94 cm

and circumference of the circle

= $2\pi r$

= $2\times \frac{22}{7}\times 14$

= 88 cm

Hence, required sum

=94+88=182 cm

Incorrect

The area of the circle= $\pi r^{2}$

= $\frac{22}{7}\times 14\times 14$

=616 sq cm

Let the length and breadth of the rectangle be x cm
and y cm respectively.
Then,

$x\times y$ =1166-616

$\Rightarrow 25\times y$ =550

$\Rightarrow y=\frac{550}{25}$ =22 cm

$\therefore$ Perimeter of the rectangle

=2(x+y)

=2(25+22)

=94 cm

and circumference of the circle

= $2\pi r$

= $2\times \frac{22}{7}\times 14$

= 88 cm

Hence, required sum

=94+88=182 cm
Question 25 of 50

25. Question
1 points
Smallest angle of a triangle.is equal to two-third the smallest angle of a quadrilateral. The ratio between the, angles of the quadrilateral is 3 : 4 : 5 : 6. Largest angle of the
triangle is twice its smallest angle. What is the sum of second largest angle of the triangle and largest angle of the quadrilateral ?
- 1. $160^{\circ}$
- 2. $180^{\circ}$
- 3. $190^{\circ}$
- 4. $170^{\circ}$
- 5. None of these
Correct

Let the angles of the quadrilateral be 3 x,4 x,5 x and 6 x respectively.
Then,

3 x + 4 x + 5 x + 6 x= $360^{\circ}$

$\Rightarrow 18 x=360^{\circ}$

$\Rightarrow x= 20^{\circ}$

$\therefore$ Smallest angle of the triangle

= $3\times 20\times \frac{2}{3}$

= $40^{\circ}$

$\therefore$ Largest angle of the triangle

= $40^{\circ}\times$ 2

= $80^{\circ}$

$\therefore$ Second largest angle of the triangle

= $180^{\circ}-(40^{\circ}+80^{\circ})$

= $60^{\circ}$

and the largest angle of the quadrilateral

=6 x

= $6\times 20^{\circ}$

=120 $^{\circ}$

Hence, required sum

= $60^{\circ}$ + $12^{\circ}$

= $180^{\circ}$

Incorrect

Let the angles of the quadrilateral be 3 x,4 x,5 x and 6 x respectively.
Then,

3 x + 4 x + 5 x + 6 x= $360^{\circ}$

$\Rightarrow 18 x=360^{\circ}$

$\Rightarrow x= 20^{\circ}$

$\therefore$ Smallest angle of the triangle

= $3\times 20\times \frac{2}{3}$

= $40^{\circ}$

$\therefore$ Largest angle of the triangle

= $40^{\circ}\times$ 2

= $80^{\circ}$

$\therefore$ Second largest angle of the triangle

= $180^{\circ}-(40^{\circ}+80^{\circ})$

= $60^{\circ}$

and the largest angle of the quadrilateral

=6 x

= $6\times 20^{\circ}$

=120 $^{\circ}$

Hence, required sum

= $60^{\circ}$ + $12^{\circ}$

= $180^{\circ}$
Question 26 of 50

26. Question
1 points
A 320 m long train moving with an average speed of 120 km/h crosses a platform in 24 s. A man crosses the same platform in 4 min. What is the speed of man in m/s?
- 1. 2.4
- 2. 1.5
- 3. 1.6
- 4. 2.0
- 5. None of these
Correct

Speed of train

= 120 km/h

= $120\times \frac{5}{18}$ m/s

= $\frac{100}{3}$

Suppose the length of the platform = x m
Then,

$\frac{x+320}{100}$ =24

$\Rightarrow 3(x+320)=100\times 24$

$\Rightarrow$ x+320=800

$\Rightarrow$ x=800-320=480 m

Hence, speed of man

= $\frac{480}{4\times 60 }$ m/s

=2 m/s

Incorrect

Speed of train

= 120 km/h

= $120\times \frac{5}{18}$ m/s

= $\frac{100}{3}$

Suppose the length of the platform = x m
Then,

$\frac{x+320}{100}$ =24

$\Rightarrow 3(x+320)=100\times 24$

$\Rightarrow$ x+320=800

$\Rightarrow$ x=800-320=480 m

Hence, speed of man

= $\frac{480}{4\times 60 }$ m/s

=2 m/s
Question 27 of 50

27. Question
1 points
The simple interest accrued on a sum of certain principal is Rs. 7200 in six years at the rate of 12 p.c.p.a. What would be the compound interest accrued on that principal at the
rate of 5 p.c.p.a. in 2 yr?
- 1. Rs.1020
- 2. Rs.1055
- 3. Rs.1050
- 4. Rs.1025
- 5. None of these
Correct

Suppose the principal amount= Rs.x

Then,

7200= $\frac{x\times 12\times 6}{100}$

$\Rightarrow x=\frac{7200\times 100}{12\times 6}$

$\Rightarrow$ x=Rs.10000

$\therefore$ amount interest

= $10000(1+\frac{5}{100})^{2}-10000$

= $10000(\frac{21}{20})^{2}-10000$

= $10000\times \frac{441}{400}-10000$

= $\frac{410000-4000000}{400}$

= $\frac{410000}{400}$

= Rs.1025.

Incorrect

Suppose the principal amount= Rs.x

Then,

7200= $\frac{x\times 12\times 6}{100}$

$\Rightarrow x=\frac{7200\times 100}{12\times 6}$

$\Rightarrow$ x=Rs.10000

$\therefore$ amount interest

= $10000(1+\frac{5}{100})^{2}-10000$

= $10000(\frac{21}{20})^{2}-10000$

= $10000\times \frac{441}{400}-10000$

= $\frac{410000-4000000}{400}$

= $\frac{410000}{400}$

= Rs.1025.
Question 28 of 50

28. Question
1 points
Sum of square of first number and cube of second numbers 568 together. Also square of the second number is 15 less than square of 8. What is the value of three-fifth of the
first number? (assuming both the numbers are positive)
- 1. 18
- 2. 8
- 3. 9
- 4. 16
- 5. None of these
Correct

Suppose the first number=x

and second number=y

Then,

$8^{2}-y^{2}=15$

$\Rightarrow 64-y^{2}=15$

$\Rightarrow y^{2}=49$

$\Rightarrow y=\sqrt{49}=7$

Again,

$x^{2}+y^{3}=568$

$\Rightarrow x^{2}+7^{3}=568(on putting value y=7)$

$\Rightarrow x^{2}+343=568$

$\Rightarrow x^{2}=225$

$\Rightarrow x=\sqrt{225}=15$

Hence, three fifth of the first number

= $\frac{3}{5}\times 15$

=9.

Incorrect

Suppose the first number=x

and second number=y

Then,

$8^{2}-y^{2}=15$

$\Rightarrow 64-y^{2}=15$

$\Rightarrow y^{2}=49$

$\Rightarrow y=\sqrt{49}=7$

Again,

$x^{2}+y^{3}=568$

$\Rightarrow x^{2}+7^{3}=568(on putting value y=7)$

$\Rightarrow x^{2}+343=568$

$\Rightarrow x^{2}=225$

$\Rightarrow x=\sqrt{225}=15$

Hence, three fifth of the first number

= $\frac{3}{5}\times 15$

=9.
Question 29 of 50

29. Question
1 points
The sum of 8 consecutive odd numbers is 656. Also average of four consecutive even number is 87. What is the sum of the smallest odd number and second largest
even number?
- 1. 165
- 2. 175
- 3. 163
- 4. cannot be determined
- 5. None of these
Correct

Let the 8 consecutive odd numbers be

x, x+2, x+4, x+6, x+8, x+10, x+12, and x+14 respectively.

Then,

x+(x+2)+(x+6)+(x+8)+(x+10)+(x+12)+(x+14)=656

$\Rightarrow$ 8 x+56=656

$\Rightarrow x=\frac{600}{8}$ =75

Let the four consecutive even numbers be

x,x+2,x+4,x+6,x+8,x+10,x+12,x,x+2,x+4 and x+6 respectively.

Then,

x+(x+2)+(x+6)+(x+8)

$\Rightarrow 4 x+12=4\times 87$

$\Rightarrow$ 4 x=348-12

$\Rightarrow x=\frac{336}{4}=84$

$\therefore$ Second largest even number

=x+4

=84+4

=88

Hence,required sum

=75+88

=163.

Incorrect

Let the 8 consecutive odd numbers be

x, x+2, x+4, x+6, x+8, x+10, x+12, and x+14 respectively.

Then,

x+(x+2)+(x+6)+(x+8)+(x+10)+(x+12)+(x+14)=656

$\Rightarrow$ 8 x+56=656

$\Rightarrow x=\frac{600}{8}$ =75

Let the four consecutive even numbers be

x,x+2,x+4,x+6,x+8,x+10,x+12,x,x+2,x+4 and x+6 respectively.

Then,

x+(x+2)+(x+6)+(x+8)

$\Rightarrow 4 x+12=4\times 87$

$\Rightarrow$ 4 x=348-12

$\Rightarrow x=\frac{336}{4}=84$

$\therefore$ Second largest even number

=x+4

=84+4

=88

Hence,required sum

=75+88

=163.
Question 30 of 50

30. Question
1 points
Seema purchased an item for Rs.9600 and sold it for loss of 5 per cent. From that money she purchased another item and sold it for gain of 5 per cent. What is her overall gain/loss?
- 1. Loss of Rs.36
- 2. Profit of Rs.24
- 3. Loss of Rs.54
- 4. Profit of Rs.36
- 5. None of these
Correct

Cost price of the item =Rs.9600
Selling price of the item
=9600 $\times \frac{95}{100}\times \frac{105}{100}$
=Rs.9576
Hence, required loss
=9600-9576
=Rs.24

Incorrect

Cost price of the item =Rs.9600
Selling price of the item
=9600 $\times \frac{95}{100}\times \frac{105}{100}$
=Rs.9576
Hence, required loss
=9600-9576
=Rs.24
Question 31 of 50

31. Question
1 points
Study the table carefully ta answer the questions that follow :

What was the total number of failed candidates from school-C in the year 2008 and the number of candidates who appeared in the exam from school-D in the year 2006 ?
- 1. 335
- 2. 325
- 3. 322
- 4. 332
- 5. None of these
Correct

Required number of candidates
=(354-258)+235
=96+235
=331

Incorrect

Required number of candidates
=(354-258)+235
=96+235
=331
Question 32 of 50

32. Question
1 points
Study the table carefully to answer the questions that follow :

In which year was the difference between the number of candidates who appeared and passed in the exam from school-B second lowest?
- 1. 2004
- 2. 2005
- 3. 2006
- 4. 2007
- 5. 2008
Correct

In year 2004 = 445 - 354 = 91
In year 2005 = 545 - 435 = 110
In year 2006 = 664- 454=210
In year 2007 = 345 -144 = 201
In year 2008 = 584 - 354 = 23O
Hence, in year 2005 the difference between the
appeared and passed candidates from school-B was
second lowest.

Incorrect

In year 2004 = 445 - 354 = 91
In year 2005 = 545 - 435 = 110
In year 2006 = 664- 454=210
In year 2007 = 345 -144 = 201
In year 2008 = 584 - 354 = 23O
Hence, in year 2005 the difference between the
appeared and passed candidates from school-B was
second lowest.
Question 33 of 50

33. Question
1 points
Study the table carefully ta answer the questions that follow :

What was the respective ratio between the number of candidates who appeared from school-C in the year 2006 and the number of candidates who passed in the exam
from school-D in the Year 2009 ?
- 1. 11 : 4
- 2. 11 : 5
- 3. 5 : 11
- 4. 9 : 11
- 5. None of these
Correct

Required respective ratio
= 693 : 252
= 11 : 4

Incorrect

Required respective ratio
= 693 : 252
= 11 : 4
Question 34 of 50

34. Question
1 points
Study the table carefully to answer the question that follow :

Number of candidates who passed in the exam from school B in the year 2005 was approximately what per cent of number of candidates who appeared from
school-A in the Year 2008?
- 1. 76
- 2. 87
- 3. 90
- 4. 84
- 5. 80
Correct

Required percentage
= $\frac{435}{546}\times 100$
=79.67%
=80%(Approx.)

Incorrect

Required percentage
= $\frac{435}{546}\times 100$
=79.67%
=80%(Approx.)
Question 35 of 50

35. Question
1 points
Study the table carefully to answer the questions that follow :

What was the approximate percent increase in the number of candidates who passed in the exam from school-A in the year 2009 as compared to the previous year?
- 1. 22
- 2. 39
- 3. 26
- 4. 30
- 5. 34
Correct

Required percentage

= $\frac{435-346}{346}\times 100$

= $\frac{89}{346}\times 100$

=25.72%

=26%(Approx.)

Incorrect

Required percentage

= $\frac{435-346}{346}\times 100$

= $\frac{89}{346}\times 100$

=25.72%

=26%(Approx.)
Question 36 of 50

36. Question
1 points
Study the fol]owing graph carefully to answer the questions that follow :

What was the average number of soldiers recruited in the Navy overall the Years together?
- 1. 25000
- 2. 24000
- 3. 2400
- 4. 28000
- 5. None of these
Correct

Required average number of soldiers
= $\frac{(5+10+15+30+40+50)}{6}\times 100$
= $\frac{150}{6}\times 100$
=25000

Incorrect

Required average number of soldiers
= $\frac{(5+10+15+30+40+50)}{6}\times 100$
= $\frac{150}{6}\times 100$
=25000
Question 37 of 50

37. Question
1 points
Study the following graph carefully to answer the questions that follow :

Number of soldiers recruited in Navy in the year 2009 was what percentage of soldiers recruited in Army in the year 2006 ?
- 1. 140
- 2. 150
- 3. 160
- 4. 180
- 5. None of thes
Correct

Required percentage
= $\frac{40}{25}\times 100$
=160%

Incorrect

Required percentage
= $\frac{40}{25}\times 100$
=160%
Question 38 of 50

38. Question
1 points
Study the following graph carefully to answer the questions that follow :

If 30 per cent of soldiers recruited in Airforce in the year 2010 was female then what is the number of males recruited in Air force in that year?
- 1. 63000
- 2. 6300
- 3. 61000
- 4. 6100
- 5. None of these
Correct

Required numbers of males
= $90\times \frac{(100-30)}{100}\times 1000$
= $90\times \frac{70}{100}\times 1000$
=63000

Incorrect

Required numbers of males
= $90\times \frac{(100-30)}{100}\times 1000$
= $90\times \frac{70}{100}\times 1000$
=63000
Question 39 of 50

39. Question
1 points
Study the following graph carefully to answer the questions that follow :

What was the respective ratio between the number of soldiers recruited for Airforce in the year 2005 and the number soldiers recruited in Army in the year 2009 ?
- 1. 2 : 15
- 2. 5 : 13
- 3. 2 : 17
- 4. 15 : 4
- 5. None of these
Correct

Required respective ratio

= 10 : 75

= 2 : 15

Incorrect

Required respective ratio

= 10 : 75

= 2 : 15
Question 40 of 50

40. Question
1 points
Study the following graph carefully to answer the questions that follow :

What was approximate percentage decrease in number of soldiers recruited in Army in the year 2008 as compared to the previous year ?
- 1. 20
- 2. 23
- 3. 38
- 4. 30
- 5. 33
Correct

Required decrease percentage
= $\frac{45-30}{45}\times 100$
= $\frac{15}{45}\times 100$
=33.33%
=33%(Approx.)

Incorrect

Required decrease percentage
= $\frac{45-30}{45}\times 100$
= $\frac{15}{45}\times 100$
=33.33%
=33%(Approx.)
Question 41 of 50

41. Question
1 points
Study the following pie-chart carefully to answer these questions :

What is the value of half of the difference between the number of students in MBA and MBBS ?
- 1. 800
- 2. 1600
- 3. 1300
- 4. 650
- 5. None of these
Correct

Total number of MBA students
= $6500\times \frac{26}{100}$
=1690
Total number of MBBS students
= $6500\times \frac{6}{100}$
=390
Hence, required difference
= $\frac{1}{2}(1690-390)$
= $\frac{1}{2}\times 1300$
=650

Incorrect

Total number of MBA students
= $6500\times \frac{26}{100}$
=1690
Total number of MBBS students
= $6500\times \frac{6}{100}$
=390
Hence, required difference
= $\frac{1}{2}(1690-390)$
= $\frac{1}{2}\times 1300$
=650
Question 42 of 50

42. Question
1 points
Study the following pie-chart carefully to answer these questions :

How much more percentage (approximately) of students are in MBA as compared to students in B.Ed.?
- 1. 49
- 2. 53
- 3. 59
- 4. 41
- 5. 44
Correct

Total number of B.ED. students
= $6500\times \frac{18}{100}$
=1700
Total number of MBA students
= $6500\times \frac{26}{100}$
=1690
Hence, required percentage
= $\frac{1690-1170}{1170}\times 100$
= $\frac{520}{1170}\times 100$
=44.44%
=44%(Approx.)

Incorrect

Total number of B.ED. students
= $6500\times \frac{18}{100}$
=1700
Total number of MBA students
= $6500\times \frac{26}{100}$
=1690
Hence, required percentage
= $\frac{1690-1170}{1170}\times 100$
= $\frac{520}{1170}\times 100$
=44.44%
=44%(Approx.)
Question 43 of 50

43. Question
1 points
Study the following pie-chart carefully to answer these questions :

What is the total number of students in B.Ed., Pharmacy and MBBS together?
- 1. 2465
- 2. 2565
- 3. 2405
- 4. 2504
- 5. None of these
Correct

Required number of students
= $6500\times \frac{18}{100}+6500\times \frac{13}{100}+6500\times \frac{6}{100}$
=1170+845+390
=2405

Incorrect

Required number of students
= $6500\times \frac{18}{100}+6500\times \frac{13}{100}+6500\times \frac{6}{100}$
=1170+845+390
=2405
Question 44 of 50

44. Question
1 points
Study the following pie-chart carefully to answer these questions :

What is the respective ratio between the number of students in Pharmacy and the number of students in B.Tech?
- 1. 11 : 13
- 2. 13 : 6
- 3. 13 : 7
- 4. 6 : 13
- 5. None of these
Correct

Required respective ratio
= $6500\times \frac{13}{100}:6500\times \frac{7}{100}$
=845 : 455
=13 : 7

Incorrect

Required respective ratio
= $6500\times \frac{13}{100}:6500\times \frac{7}{100}$
=845 : 455
=13 : 7
Question 45 of 50

45. Question
1 points
Study the following pie-chart carefully to answer these questions :

Number of students in B.Sc. is approximately what percentage of the number of students in B.Ed.?
- 1. 167
- 2. 162
- 3. 157
- 4. 153
- 5. 150
Correct

Total number of B.Sc students
= $6500\times \frac{30}{100}$
=1950
Total number of B.Ed students
= $6500\times \frac{18}{100}$
=1170
Hence, required percentage
= $\frac{1950}{1170}\times 100$
=166.66%
=167%(Approx.)

Incorrect

Total number of B.Sc students
= $6500\times \frac{30}{100}$
=1950
Total number of B.Ed students
= $6500\times \frac{18}{100}$
=1170
Hence, required percentage
= $\frac{1950}{1170}\times 100$
=166.66%
=167%(Approx.)
Question 46 of 50

46. Question
1 points
Study the following table carefully to answer the questions that follow.

What was the average of the earning of Person-B in the year 2006, C in the year 2008 and E in the year 2005 together?
- 1. Rs.3.62 lac
- 2. Rs.2.64 lac
- 3. Rs.3.64 lac
- 4. Rs.10.86 lac
- 5. None of these
Correct

Required average earning
=Rs. $\frac{3.34+5.83+1.69}{3}$ lac
=RS. $\frac{10.86}{3}$ lac
=Rs.3.62 lac

Incorrect

Required average earning
=Rs. $\frac{3.34+5.83+1.69}{3}$ lac
=RS. $\frac{10.86}{3}$ lac
=Rs.3.62 lac
Question 47 of 50

47. Question
1 points
Study the following table carefully to answer the questions that follow :

What was the respective ratio between the amount earned by Person-B in the year 2007 and Person-D in the year 2010 ?
- 1. 32 : 107
- 2. 31 : 105
- 3. 29 : 107
- 4. 32 : 105
- 5. None of these
Correct

Required respective ratio
= 2.79 : 9.45
= 279 : 945
= 31 : 105

Incorrect

Required respective ratio
= 2.79 : 9.45
= 279 : 945
= 31 : 105
Question 48 of 50

48. Question
1 points
Study the following table carefully to answer the questions that follow.

What is the approximate per cent increase in the amount earned by Person-D in the year 2010 as compared to the previous year?
- 1. 7
- 2. 21
- 3. 18
- 4. 15
- 5. 12
Correct

Required increase percentage
= $\frac{9.45-8.42}{8.42}\times 100$
= $\frac{1.03}{8.42}\times 100$
=12.33%
=12%(Approx.)

Incorrect

Required increase percentage
= $\frac{9.45-8.42}{8.42}\times 100$
= $\frac{1.03}{8.42}\times 100$
=12.33%
=12%(Approx.)
Question 49 of 50

49. Question
1 points
Study the following table carefully to answer the questions that follow.

Whose earning increased consistently from the year 2005 to the year 2010?
- 1. A
- 2. B
- 3. C
- 4. D
- 5. E
Correct

D's earning increased consistently from the year 2005 to the Year 2010.

Incorrect

D's earning increased consistently from the year 2005 to the Year 2010.
Question 50 of 50

50. Question
1 points
Study the following table carefully to answer the questions that follow.

Total amount earned by Person-A in the year 2006 and Person-C in the year 2010 together was approximately what per cent of the amount earned by Person-E in the year 2009?
- 1. 151
- 2. 155
- 3. 168
- 4. 174
- 5. 162
Correct

Required percentage
= $\frac{1.44+7.84}{5.53}\times 100$
= $\frac{9.28}{5.53}\times 100$
=167.81%
=168%(Approx.)

Incorrect

Required percentage
= $\frac{1.44+7.84}{5.53}\times 100$
= $\frac{9.28}{5.53}\times 100$
=167.81%
=168%(Approx.)

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