SSC CHSL Simplification Questions Solved Problems with Detailed Solutions [Free PDF]

Concept

Calculation

⇒ 441 ÷ $[270÷(\frac{3}{7}\times 35)+(17\times 3)-(17/2-5/2)]$

⇒ 441 ÷ [270 ÷ (3 × 5) + 51 - $(\frac{12}{2})]$

⇒ 441 ÷ [ 270 ÷ 15 + 51 - 6]

⇒ 441 ÷ [ 18 + 51 - 6]

⇒ 441 ÷ 63

⇒  7

∴ The value of $441÷[270÷\frac{3}{7}of35+(17÷\frac{1}{3})-(8\frac{1}{2}-\frac{5}{2})]$ is 7

Concept

BODMAS rule applies here.

First solve bracket values, after it solves the values related with 'Of', Then Divided value, afterward Multiply values, Then addition and subtractions values.

Calculation

$\frac{3\frac{2}{3}-5\frac{7}{9}÷\frac{1}{3}\times 1\frac{2}{13}}{\frac{2}{3}of1\frac{1}{3}÷\frac{1}{3}}$

⇒ $\frac{[\frac{11}{3}-\frac{52}{9}\times 3\times \frac{15}{13}]}{\frac{2}{3}\times \frac{4}{3}\times 3}$

⇒ $\frac{[\frac{11}{3}-\frac{20}{1}]}{\frac{8}{9}\times 3}$

⇒ $\frac{11-60}{3}$ × $\frac{3}{8}$

⇒  $\frac{-49}{8}$

⇒ $-6\frac{1}{8}$

∴ The value of $\frac{3\frac{2}{3}-5\frac{7}{9}÷\frac{1}{3}\times 1\frac{2}{13}}{\frac{2}{3}of1\frac{1}{3}÷\frac{1}{3}}$ is $-6\frac{1}{8}$

Concept Used:       

Follow the BODMAS rule according to the table given below:

Calculation:

$42÷10-(\frac{7}{4}\times \frac{1}{3})of\frac{6}{5}+\frac{16}{3}\times \frac{9}{10}-(\frac{1}{6}÷\frac{1}{36})$

⇒ 42 ÷ 10 – 7/12 of 6/5 + 16/3 × 9/10 – (1/6 ÷ 1/36)

⇒ 42 ÷ 10 – 7/12 of 6/5 + 16/3 × 9/10 – (1/6 × 36/1)

⇒ 42 ÷ 10 – 7/12 of 6/5 + 16/3 × 9/10 – 6

⇒ 42 ÷ 10 – 7/10 + 16/3 × 9/10 – 6

⇒ 42/10 – 7/10 + 24/5 – 6

⇒ (42 – 7 + 48)/10 – 6

⇒ (83/10 – 6)

⇒ (83 – 60)/10

⇒ 23/10

⇒ 2.3

∴ The required value is 2.3

Given:

$18\frac{2}{11}÷[\frac{7}{33}+\{\frac{29}{6}-\frac{3}{11}\times (5\frac{1}{6}-3\frac{1}{3})\}]$

Concept used:

Calculation:

$18\frac{2}{11}÷[\frac{7}{33}+\{\frac{29}{6}-\frac{3}{11}\times (5\frac{1}{6}-3\frac{1}{3})\}]$

⇒ 200/11 ÷ [7/33 + {29/6 – 3/11 × (31/6 – 10/3)}]

⇒ 200/11 ÷ [7/33 + {29/6 – 3/11 × 11/6}]

⇒ 200/11 ÷ [7/33 + {29/6 – 1/2}]

⇒ 200/11 ÷ [7/33 + 13/3]

⇒ 200/11 ÷ 150/33

⇒ 200/11 × 33/150

⇒ 4

∴ Required answer is 4

Calculations :

25% of 480 + 30% of 500 + x% of 90 = 35% of 900

⇒ (25/100) of 480 + (30/100) of 500 + (x/100) of 90 = (35/100) of 900

⇒ 120 + 150 + (9x/10) = 315

⇒ (9x/10) = 45

⇒ x = 50 

∴ The correct choice will be option 2.

Concept Used:

Follow the BODMAS rule according to the table given below:

Calculation:

$(1\frac{1}{8}÷\frac{3}{4})÷[\left(\frac{1}{2}of1\frac{1}{2}\right)÷(3\frac{1}{2}-1\frac{1}{4})of\frac{3}{4}-\frac{3}{4}÷2\frac{1}{4}]of\frac{3}{4}$

⇒ (9/8 ÷ 3/4) ÷ [(1/2 of 3/2) ÷ (7/2 – 5/4) of 3/4 – 3/4 ÷ 9/4] of 3/4

⇒ (9/8 × 4/3) ÷ [(3/4) ÷ (9/4) of 3/4 – 1/3] of 3/4

⇒ (3/2) ÷ [(3/4) ÷ 27/16 – 1/3] of 3/4

⇒ (3/2) ÷ [3/4 × 16/27 – 1/3] of 3/4

⇒ (3/2) ÷ [4/9 – 1/3] of 3/4

⇒ (3/2) ÷ [1/9] of 3/4

⇒ (3/2) ÷ 1/12

⇒ (3/2) × 12

⇒ 18

∴ The required value is 18

Given:

 [7 + 7 × ( 7 + 7 ÷ 7)] + 7 ÷ 7. 

Concept used:

Calculation:

⇒ [7 + 7 × ( 7 + 7 ÷ 7)] + 7 ÷ 7.

⇒ [7 + 7 × (7 + 1)] + 1

⇒ 63 + 1 = 64 

∴ The value is 64.

Given:

$8-[8-(5+8)-\{8-(8-5+8)\}+10]$

Concept used:

Calculation:

8 - [8 - 13 - {8 - 11} + 10]

⇒ 8 - [8 - 13 +3 + 10]

⇒ 8 - 8

∴ This value is 0

Concept Used:

Calculations:

Considering the given equation,

10 – [121 ÷ (11 × 11) – (- 4) – {3 – (8 – 1)}] = ?

⇒ 10 – [121 ÷ (121) + 4 – {3 – 7}] = ?

⇒ 10 – [121 ÷ (121) + 4 – {-4}] = ?

⇒ 10 – [121 ÷ (121) + 4 + 4] = ?

⇒ 10 – [121/121 + 4 + 4] = ?

⇒ 10 – [1 + 4 + 4] = ?

⇒ 10 – 9 = ?

⇒ 1 = ?

∴ The value of ? is 1.

Concept used:

Follow BODMAS rule to solve this question, as per the order given below,

Now,

Calculation:

$[5\frac{4}{9}÷{(\frac{11}{4}-\frac{13}{6})}^2]÷[7\frac{3}{11}of8\frac{4}{5}÷1\frac{5}{7}-\frac{4}{3}{]}^2$

$\Rightarrow \left[\frac{49}{9}÷{\left(\frac{33-26}{12}\right)}^2\right]÷{\left[\frac{80}{11}of\frac{44}{5}÷\frac{12}{7}-\frac{4}{3}\right]}^2$

$\Rightarrow \left[\frac{49}{9}\times \frac{144}{49}\right]÷{\left[64\times \frac{7}{12}-\frac{4}{3}\right]}^2$

⇒ 16 ÷ [112/3 – 4/3]²

⇒ 16 ÷ 36²

⇒ 16/1296

∴ 1/81

Concept used:

Calculation:

3 × 6 ÷ 4 of 6 – 6 ÷ 2 × (4 - 6) + 4 – 2 × 3 ÷ 6 of 1/3

⇒ 3 × 6 ÷ 4 of 6 – 6 ÷ 2 × -2 + 4 – 2 × 3 ÷ 6 of 1/3

⇒ 3 × 6 ÷ 24 – 6 ÷ 2 × -2 + 4 – 2 × 3 ÷ 2

⇒ 3 × 1/4 – 3 × -2 + 4 – 2 × 3/2

⇒ 3/4 + 6 + 4 – 3

⇒ 43/4 – 3

⇒ 31/4

⇒ 7$\frac{3}{4}$

Calculation:

⇒ $\frac{1.0025+6.25\times {10}^{-6}}{0.0025+0.95}$

$\Rightarrow \frac{1.0025+0.00000625}{0.0025+0.95}$

$\Rightarrow \frac{1.00250625}{0.9525}$

⇒ 1.0525

∴ The correct answer is 1.0525.

Follow BODMAS rule to solve this question, as per the order given below,

Step - 1 - Parts of an equation enclosed in 'Brackets' must be solved first,

Step - 2 - Any mathematical 'Of' or 'Exponent' must be solved next,

Step - 3 - Next, the parts of the equation that contain 'Division' and 'Multiplication' are calculated,

Step - 4 - Last but not least, the parts of the equation that contain 'Addition' and 'Subtraction' should be calculated.

Now, given expression:

$\Rightarrow \frac{1}{2}of\frac{8}{5}÷\{2\frac{1}{5}-(\frac{5}{16}+\frac{3}{5}\times 1\frac{7}{8}÷\frac{2}{3})\}$

$\Rightarrow \frac{1}{2}\times \frac{8}{5}÷\{\frac{11}{5}-(\frac{5}{16}+\frac{3}{5}\times \frac{15}{8}÷\frac{2}{3})\}$

$\Rightarrow \frac{4}{5}÷\{\frac{11}{5}-(\frac{5}{16}+\frac{3}{5}\times \frac{15}{8}\times \frac{3}{2})\}$

$\Rightarrow \frac{4}{5}÷\{\frac{11}{5}-(\frac{5}{16}+\frac{27}{16})\}$

$\Rightarrow \frac{4}{5}÷\{\frac{11}{5}-\frac{32}{16}\}$

$\Rightarrow \frac{4}{5}÷\frac{1}{5}$

$\Rightarrow \frac{4}{5}\times \frac{5}{1}=4$

Calculation:

$\sqrt{18}-\frac{1}{\sqrt{2}}$

⇒ $3\sqrt{2}-\frac{1}{\sqrt{2}}$

⇒ $\frac{\left(6-1\right)}{\sqrt{2}}$

⇒ $\frac{5}{\sqrt{2}}$

∴ simplified value of  $\sqrt{18}-\frac{1}{\sqrt{2}}$ is $\frac{5}{\sqrt{2}}$

Calculation:

$\frac{1}{\sqrt{25}-\sqrt{24}}-\frac{1}{\sqrt{24}-\sqrt{23}}+\frac{1}{\sqrt{23}-\sqrt{22}}-\frac{1}{\sqrt{22}-\sqrt{21}}+\frac{1}{\sqrt{21}-\sqrt{20}},$

$\Rightarrow \frac{25-24}{\sqrt{25}-\sqrt{24}}-\frac{24-23}{\sqrt{24}-\sqrt{23}}+\frac{23-22}{\sqrt{23}-\sqrt{22}}-\frac{22-21}{\sqrt{22}-\sqrt{21}}+\frac{21-20}{\sqrt{21}-\sqrt{20}},$

$\frac{\left(\sqrt{25}-\sqrt{24}\right)\left(\sqrt{25}+\sqrt{24}\right)}{\sqrt{25}-\sqrt{24}}-\frac{\left(\sqrt{24}-\sqrt{23}\right)\left(\sqrt{24}+\sqrt{23}\right)}{\sqrt{24}-\sqrt{23}}+\frac{\left(\sqrt{23}-\sqrt{22}\right)\left(\sqrt{23}+\sqrt{22}\right)}{\sqrt{23}-\sqrt{22}}-\frac{\left(\sqrt{22}-\sqrt{21}\right)\left(\sqrt{22}+\sqrt{21}\right)}{\sqrt{22}-\sqrt{21}}+\frac{\left(\sqrt{21}-\sqrt{20}\right)\left(\sqrt{21}+\sqrt{20}\right)}{\sqrt{21}-\sqrt{20}}$

⇒ (√25 + √24) - (√24 + √23) + (√23 + √22) - (√22 + √21) + (√21 + √20),

⇒ √25 + √20,

⇒ 5 + 2√5