Simplification for IBPS clerk test 9-Solved questions with detailed explanations for SSC, IBPS PO, SBI PO, RRB, RBI, CSE, CGL, CHSL, CDA, CSE, UPSC

6)

$\frac{\frac{1}{5} + 999 \frac{494}{495} \times 99}{4} = ?$

A)
25000
B)
24225
C)
24800
D)
24750
E)
None of these

Answer: Option D

Explanation

From the above question

we apply VBODMAS rule,

$= \frac{\frac{1}{5} + 999 \frac{494}{495} \times 99}{4} = ?$

Convert the mixed number to an improper fraction

$= \frac{\frac{1}{5} + \frac{494999}{495} \times 99}{4} = ?$

Reduce the numbers

$= \frac{\frac{1}{5} + \frac{494999}{5}}{4} = ? a d d t h e f r a c t i o n s = \frac{99000}{4} (r e d u c e t h e f r a c t i o n) = 24750$

Hence, option D is correct.

Discussion

7)

$\frac{{(m - n)}^{3} + {(n - r)}^{3} + {(r - m)}^{3}}{6 (m - n) (n - r) (r - m)} = ?$

A)
$\frac{1}{2}$
B)
$\frac{1}{3}$
C)
$\frac{1}{5}$
D)
$\frac{1}{6}$
E)
None of these

Answer: Option A

Explanation

In this above question,

we know that,

$= a^{3} + b^{3} + c^{3} - 3 a b c = (a + b + c) \times \frac{1}{2} [{(a - b)}^{2} + {(b - c)}^{2} + {(a - c)}^{2}] - - - (1)$

Put the value of (a+b+c)=0 from equation (1)

$= a^{3} + b^{3} + c^{3} - 3 a b c = 0 \times \frac{1}{2} [{(a - b)}^{2} + {(b - c)}^{2} + {(a - c)}^{2}] = a^{3} + b^{3} + c^{3} - 3 a b c = 0 = a^{3} + b^{3} + c^{3} = 3 a b c$

∴ Given expression in the question is

$= \frac{{(m - n)}^{3} + {(n - r)}^{3} + {(r - m)}^{3}}{6 (m - n) (n - r) (r - m)} = \frac{(a^{3} + b^{3} + c^{3})}{6 a b c} = \frac{3 a b c}{6 a b c} = \frac{1}{2}$

Hence, Option A is correct.

Discussion

8)

99÷9÷3=?

A)
$3 \frac{2}{3}$
B)
33
C)
3
D)
1
E)
None of these

Answer: Option A

Explanation

From the above question,

we apply VBODMAS rule,

= 99÷9÷3=?

=11÷3=?

= $\frac{11}{3}$

Alternative form = $3 \frac{2}{3}$ or 3.6

Hence, Option A is correct.

Discussion

9)

$13 \frac{3}{7} + 18 \frac{1}{14} + 8 \frac{3}{4} = ?$

A)
$39 \frac{1}{2}$
B)
$39 \frac{2}{14}$
C)
$40 \frac{1}{7}$
D)
$40 \frac{1}{7}$
E)
None of the above

Answer: Option D

Explanation

From the above question

we apply VBODMAS rule,

$= 13 \frac{3}{7} + 18 \frac{1}{14} + 8 \frac{3}{4}$

(Add the whole and fraction parts separately)

$= (13 + 18 + 8) + (\frac{3}{7} + \frac{1}{14} + \frac{3}{4}) = 39 + \frac{5}{4}$

(Convert the improper fraction to a mixed number)

$= 39 + 1 \frac{1}{4}$

(Separate the mixed number)

$= 39 + 1 + \frac{1}{4} = 40 + \frac{1}{4}$

Hence, option D is correct.

Discussion

10)

$756 o f \frac{2}{3} o f \frac{4}{9} o f \frac{3}{7} = ?$

A)
96
B)
108
C)
88
D)
86
E)
None of these

Answer: Option A

Explanation

Given that

$756 o f \frac{2}{3} o f \frac{4}{9} o f \frac{3}{7} = ?$

By applying VBODMAS rule, we get

$= 756 \times \frac{2}{3} \times \frac{4}{9} \times \frac{3}{7} = ?$

(Reduce the number with

greatest common divisor 3)

$= 756 \times 2 \times \frac{4}{9} \times \frac{1}{7} = ?$

$= 84 \times 2 \times 4 \times \frac{1}{7} = ?$

(Reduce the number with '9' )

= 12×2×4= ?

(Reduce the number with '7' )

= 96 =?

Hence, Option A is correct.

Discussion

Simplification-Test 9 (Complete Revision)

Answer: Option D

Answer: Option A

Answer: Option A

Answer: Option D

Answer: Option A