Simplification for IBPS clerk test 5-Solved questions with detailed explanations for SSC, IBPS PO, SBI PO, RRB, RBI, CSE, CGL, CHSL, CDA, CSE, UPSC

21)

What will come in place of question mark (?) in the following questions

${(15 \times 0.40)}^{4} \div {(1080 \times \div 30)}^{4} \times {(27 \times 8)}^{4}$

[Punjab and Sindh Bank (PO), 2011]

A)
8
B)
3
C)
12
D)
16
E)
None of these

Answer: Option B

Explanation

$\begin{array}{l} (b) Here, \\ = {(15 \times 0.40)}^{4} {÷(1080 + 30)}^{4} \times {(27 \times 3)}^{4} = {(3 \times 2)}^{4} \\ = (3 \times 2) \\ \Rightarrow {(6.00)}^{4} \div {(36)}^{4} \times {(216)}^{4} {= (6)}^{?+5} \\ \Rightarrow {(6)}^{4} {+ (6}^{2})^{4} \times {(6^{3})}^{4} {= (6)}^{?+5} \\ \Rightarrow 6^{4} \div 6^{8} \times 6^{12} {= (6)}^{?+5} \\ \Rightarrow 6^{4-8} \times 6^{12} {= 6}^{?+5} \\ \Rightarrow 6^{-4} \times 6^{12} {=6}^{?+5} \\ \Rightarrow 6^{-4+12} {=6}^{?+5} \\ \Rightarrow ?+5=8 \\ ?= 8 - 5=3 \end{array}$

Hence, Option B is correct.

Discussion

22)

$\begin{array}{l} (\sqrt{8} \times {\sqrt{8}}^{1 / 2} {) + (9)}^{1 / 2} \\ {= (?)}^{3} + \sqrt{8} - 340 \end{array}$

[Union Bank of India (PO), 2011]

A)
7
B)
19
C)
18
D)
9
E)
None of these

Answer:

Explanation

$\begin{array}{l} Here, \\ \Rightarrow (\sqrt{8} × \sqrt{8})^{\frac{1}{2}} + \sqrt{9^{\frac{1}{2}}}) = {(?)}^{3} + \sqrt{8} - 340 \\ \Rightarrow {(8)}^{\frac{1}{2}} {+ (9)}^{\frac{1}{2}} {= (?)}^{3} + \sqrt{8} -340 \\ \begin{array}{l} \Rightarrow \sqrt{8} + \sqrt{9} - \sqrt{8} {+340=(?)}^{3} \\ \Rightarrow (?)^{3} = 343 \end{array} \\ [taking cube root on both sides] \\ ? = {}^{3}{\sqrt{343}} = 7 \end{array} $

Hence, Option A is correct.

Discussion

23)

${(0.125)}^{3} \div {(0.25)}^{2} \times {(0.5)}^{2} = {(0.5)}^{7 - ?}$

[Union Bank of India (PO), 2011]

A)
2
B)
8
C)
4
D)
0
E)
None of these

Answer: Option D

Explanation

Here,

$= {(0.125)}^{3} \div {(0.25)}^{2} \times {(0.5)}^{2} = {(0.5)}^{7 - ?} = {\{{(0.5)}^{3}\}}^{3} + {(0.5)}^{2 \times 2} \times {(0.5)}^{2} = {(0.5)}^{7 - ?} = {(0.5)}^{9 - 4} \times {(0.5)}^{2} = {(0.5)}^{7 - ?} = {(0.5)}^{5} \times {(0.5)}^{2} = {(0.5)}^{7 - ?} = {(0.5)}^{7} = {(0.5)}^{7 - ?} = 7 - ? = 7 = ? = 0$

Hence, Option D is correct.

Discussion

24)

$3^{3} \times 81^{2} \div 27^{3} = {(3)}^{?}$

[Union Bank of India (PO), 2011]

A)
3
B)
4
C)
5
D)
6
E)
None of these

Answer: Option E

Explanation

$\begin{array}{l} Here, \\ \Rightarrow 3^{3} \times 81^{2} \div 27^{3} {= (3)}^{?} \\ \Rightarrow {33 X (34 )2 + (33)3 = (3)}^{?} \\ \Rightarrow 3^{3} \times 3^{8} \div 3^{9} {= (3)}^{?} \\ \Rightarrow 3^{3+8} \div 3^{9} {= (3)}^{?} \\ \Rightarrow 3^{11 - 9} {= (3)}^{?} \\ \Rightarrow ? = 2 \end{array}$

Hence, Option E is correct.

Discussion

25)

$572 \div 26 \times 12 - {200 = (2)}^{?}$

[lOB (PO), 2010]

A)
5
B)
6
C)
8
D)
7
E)
None of these

Answer: Option B

Explanation

$\begin{array}{l} Here, \\ \Rightarrow 572 \div 26 \times 12 - {200 = (2)}^{?} \\ Applying the BODMAS rule to solve, \\ we get \\ \Rightarrow 22 \times 12 - {200 = (2)}^{?} \\ \Rightarrow {264- 200 = (2)}^{?} \\ \Rightarrow {(2)}^{?} = 64 \\ {(2)}^{?} {= (2)}^{6} \\ \Rightarrow ?=6 \end{array}$

Hence, Option B is correct.

Discussion

Simplification-Test 5 (Based on Square roots and Cube roots/Surds and Indices)

Answer: Option B

Answer:

Answer: Option D

Answer: Option E

Answer: Option B