Simplification Questions for Competitive Exams

Simplification MCQ Questions for Competitive Exams. Objective Questions selected from the previous year Exam Question paper of SSC CGL, CPO, CHSL, GD, Police, Bank and UPSSSC. Type wise Question and Answer with solution and explanation.

Simplification Questions : Type wise

Type I : BODMAS (Bracket, Of, Division, Multiplication, Addition, Subtraction)

Q.1: $3 \frac 35 \times 3\frac 35 + 2 \times 3 \frac 35 \times \frac 25 + \frac 25 \times \frac 25 = ?$
a) 15
b) 16
c) 17
d) 18

Answer

Ans : b) 16
(a+b)²= a² + b² +2ab
$(3\frac 35 + \frac 25)^2 = (\frac {18+2} {5})^2 = 4^2 = 16$

Q. 2: The simplified value of the following is :
$\dfrac{4}{15} of \dfrac{5}{8} \times 6 + 15 - 10$
a) 6
b) 3
c) 5
d) 4

Answer

Ans : a) 6
$\frac{4}{15} of \frac{5}{8} \times 6 + 15 - 10$ = $\frac{1}{6} \times 6 + 15 - 10$ =1+15-10 =6

Q.3: (0.9 × 0.9 × 0.9 + 0.1 × 0.1 × 0.1) is equal to
a) 0.73
b) 0.82
c) 0.91
d) 1.00

Answer

Ans : a) 0.73
(0.9)³ + (0.1)³ = 0.729 + 0.001 = 0.730

Q.4: The value of the following is
(0.98)³ + (0.02)³ + 3 x 0.98 x 0.02 -1
a) 1.98
b) 1.09
c) 1
d) 0

Answer

Ans : d) 0
(0.98)³ + (0.02)³ + 3 x 0.98 x 0.02 -1 = (0.98)³ + (0.02)³ + 3 x 0.98 x 0.02 (0.98 + 0.02) -1
= (0.98 + 0.02)³ -1 = 1-1 =0
(a+b)³= a³ + b³ + 3ab (a+b)

Q.5: If * represent a number, than the value of * in following equation is
$5 \frac 3* \times 3 \frac 12 = 19$
a) 7
b) 4
c) 6
d) 2

Answer

Ans : a) 7
$5 \frac 3* \times 3 \frac 12 = 19$ = $\frac {5 \times * + 3} {*} \times \frac72 = 19$
= 35 * +21 = 38 *
= 3 * = 21, * = 7

Q.6: $\frac {17}{15} \times \frac {17}{15} + \frac {2}{15} - \frac {17}{15} \times \frac {4}{15}$ is equal to
a) 0
b) 1
c) 10
d) 11

Answer

Ans : b) 1
(a – b)² = a² +b² -2ab
$(\frac {17}{15} - \frac {2}{15}) ^2 = 1^2 =1$

Q.7: $\frac {1}{30} + \frac {1}{42} + \frac {1}{56} + \frac {1}{72} + \frac {1}{90} + \frac {1}{110} = ?$
a) $\sqrt {2} \frac {2}{27}$
b) $\frac 19$
c) $\frac {5}{27}$
d) $\frac{6}{55}$

Answer

Ans : d) $\frac{6}{55}$
= $\frac{1}{5 \times 6} + \frac{1}{6 \times 7} + \frac{1}{7 \times 8} + \frac{1}{8 \times 9} + \frac{1}{9 \times 10} + \frac{1}{10 \times 11}$
= $\frac{1}{5} - \frac{1}{6} + \frac{1}{6} - \frac{1}{7} + \frac{1}{7} - \frac{1}{8} + \frac{1}{8} - \frac{1}{9} + \frac{1}{9} - \frac{1}{10} + \frac{1}{10}- \frac{1}{11}$
= $\frac 15 - \frac {1}{11} = \frac {6}{55}$

Q.8: 5-[4-{3-(3-3-6)}] is equal to
a) 10
b) 6
c) 4
d) 0

Answer

Ans : a) 10
5-[4-{3-(3-3-6)}] = 5-[4-{3-(-6)}] = 5-[4-{3+6}] = 5-[4-9] =5 -[-5] =10

Type-II: Continued Fraction : Simplification Questions

Q.9: Find the value of
$1- \dfrac{1}{1 + \dfrac {2}{3 +\dfrac{4}{5}}}$
a) $\frac {12}{29}$
b) $\frac {8}{19}$
c) $\frac {48}{29}$
d) $\frac {2}{19}$

Answer

Ans : c) $\frac {48}{29}$

Q.10: If $2= x + \dfrac {1}{1+\dfrac {1}{3+\dfrac14}}$
than the value of $x$ is
a) $\frac {18}{17}$
b) $\frac {21}{17}$
c) $\frac {13}{17}$
d) $\frac {12}{17}$

Answer

Ans : b) $\frac {21}{17}$

Simplification MCQ Questions for Competitive Exams

Type – III: Square and Square Root : Simplification for Competitive Exams

Q.11: A number of boys raised ₹ 12544 for a famine fund, each boy has given as many rupees as there were boys. The number of boys was ?
a) 102
b) 112
c) 122
d) 132

Answer

Ans : b)112
Number of Boys x Amount Contributed = 12544 = 112²
Number of Boys = Amount Contributed = 112

Q.12: The sum of two numbers is 37 and the difference of their squares is 185, then the difference between the two numbers is :
a) 10
b) 4
c) 5
d) 3

Answer

Ans : c) 5
a+b=37, a²-b² = 185
a²-b² = (a-b) (a+b)
185 = (a-b) x 37
a-b = 185/37 =5

Q.13: The value of (11111)² is
a) 12344321
b)121212121
c) 123454321
d) 11244311

Answer

Ans : c) 123454321
11² =121
111² =12321
1111² = 1234321
11111² = 123454321

Q.14: The square root of following is :
$\frac {9.5 \times 0.085}{0.017 \times 0.019}$
a) 0.5
b) 5
c) 50
d) 500

Answer

Ans : c) 50
$\frac {9.5 \times 0.085}{0.017 \times 0.019} = 50 \times 50$
Square Root is 50

Q.15: $\sqrt{64} - \sqrt{36}$ is equal to :
a) -2
b) 2
c) 0
d) 1

Answer

Ans : b) 2
8-6=2

Q.16 : The sum of the squares of 3 consecutive positive numbers is 365. The sum of the numbers is :
a) 30
b) 33
c) 36
d) 45

Answer

Ans : b) 33
10²+11²+12² = 100+121+144 = 365
10+11+12 = 33

Q.17: Given that $\sqrt {24}$ is approximately equal to 4.898. Then $\sqrt \frac 83$ is nearly equal to :
a) 0.544
b) 1.333
c) 1.633
d) 2.666

Answer

Ans : c) 1.633
$\sqrt \frac 83 = \sqrt \frac {8 \times 3}{3 \times 3} = \frac {\sqrt {24}}{3} = \frac {4.898}{3} = 1.633$

Q.18: The value of $\frac {(75.8)^2 - (55.8)^2}{20}$
a) 20
b) 40
c) 121.6
d) 131.6

Answer

Ans : d) 131.6
$\frac {(75.8)^2 - (55.8)^2}{20} = \frac {(75.8 - 55.8) (75.8+55.8)}{20} = \frac {20 \times 131.6} {20} = 131.6$

Q.19: $\sqrt { \frac {0.49}{0.25}} + \sqrt{\frac {0.81}{0.36} }$ is equal to :
a) $7 \frac {9}{10}$
b) $2 \frac {9}{10}$
c) $\frac {9}{10}$
d) $9 \frac {9}{10}$

Answer

Ans : b) $2 \frac {9}{10}$
$\sqrt { \frac {49}{25}} + \sqrt{\frac {81}{36} }$
$\frac75 +\frac 96 = \frac {87}{30} = \frac {29}{10} = 2\frac{9}{10}$

Q.20: When simplified, the product is equals to :
$(2-\frac13) (2-\frac57) .............. (2-\frac {997}{999})$
a) $latex \frac {5}{999}
b) $latex \frac {5}{3}
c) $latex \frac {1001}{999}
d) $latex \frac {1001}{3}