16+ Simplification Questions for CAT with Solutions

Q1.  Simplify 4368 + 2158 – 596 –? = 3421 + 1262

Answer. 1247.

Detailed Solution. ∴  Let 4368 + 2158 – 596 – x = 3421 + 1262

⇒ x + 596 = (4368 + 2158) – (3421 + 1262)

⇒ x + 596 = 6526 – 4683 = 1843

⇒ x = 1843 – 596 = 1247.

Hence, required number = 1247.

Q2. Simplify 3456 ÷ 12 ÷ 8 =?

Answer. 36

Detailed Solution. Given expression = 3456/12 ÷ 8 = 288 ÷ 8 = 36.

Q3. Simplify 13 × 252 ÷ 42 + 170 =? + 47

Answer. 201

Detailed Solution. Let 13 × 252 ÷ 42 + 170 = x + 47. Then, 13 × 252 42 + 170 = x + 47

⇒ 13 × 6 + 170 = x + 47

⇒ x + 47 = 78 + 170 = 248

⇒ x = 248 – 47 = 201.

Hence, the required number = 201.

Q4. Simplify: (a) 460 × 15 – 5 × 20 

(b) 1 ÷ [1 + 1 ÷ {1 + 1 ÷ 1(1 + 1 ÷ 2)}] + 1

Answer.  a) 6800

b) 13/8

Detailed Solution. (a) Given expression = 6900 – 100 = 6800.

(b) Given expression = 1 ÷ [1 + 1 ÷ {1 + 1 ÷ 1(1 + (1/2)}] + 1

= 1 ÷ [1 + 1 ÷ {1 + 1 ÷ 3/2}] + 1 = 1 ÷ [1 + 1 ÷ {1 + 1 × (2/3)}] + 1

= 1 ÷ [1 + 1 ÷ [1 + {2/3}] + 1 = 1 ÷ [1 + 1 ÷ 5/3] + 1

= 1 ÷ [1 + 1 × (3/5)] + 1 = 1 ÷ [1 + 3/5] + 1 = 1 ÷ (8/5) + 1 = 1 × 5/8 + 1 = 13/8.

Q5. Find the missing numeral: (a) (? – 2763) ÷ 86 × 13 = 208

(b) 3565 ÷ 23 + 4675 ÷? = 430

Answer.  17

Detailed Solution. (a) Let (x – 2763) ÷ 86 × 13 = 208.

Then, (x-2763)/8 × 13 = 208 ⇒ (x – 2763)/86 = 208/13 = 16
⇒ (x – 2763) = 16 × 86 = 1376
 ⇒ x = 1376 + 2763 = 4139.

(b) Let 3565 ÷ 23 + 4675 ÷ x = 430.
Then, 3565/23 + 4675/x = 430 ⇒ 155 + 4675/x = 430
⇒ 4675/x = 430 - 155 = 275 ⇒ x = 4675/275 = 17

Q6. Simplify: Simplify: (a) (6 + 6 + 6 + 6) ÷ 6 / 4 + 4 + 4 + 4 ÷ 4

(b) ((2 + 3) × 5 + 3 ÷ 1/2) / (6 + 5 × 4 ÷ 4/5)

Answer. (a) = 4/13

(b) = 1

Detailed Solution. 

(a) Given expression = 24 ÷ 6 / 4 + 4 + 4 + 1
= 4/13

(b) Given expression = (5 × 5 + 3 × 2) / (6 + 5 × 4 × 5/4)
= 25 + 6 / 6 + 25
= 31/31
= 1

Q7. The length of a rectangle is increased by 30%. By what percent would the breadth have to be decreased to maintain the same area?

Answer. 37.5%.

Detailed Solution. Let the length and breadth of the rectangle be l and b units respectively.
Then, area of rectangle = (lb) sq. units.
New length = 160% of l = 8l/5 units.

Desired breadth = ^Area⁄_{New length} = ^lb⁄_{(8l ⁄5)} = ^5b⁄₈ units.

Decrease in breadth = [ b - ^5b⁄₈ ] units = ^3b⁄₈ units.

∴ Decrease% = ( ^3b⁄₈ × ¹⁄_b × 100 )% = ⁷⁵⁄₂% = 37.5%.

Q8. In measuring the sides of a rectangular plot, one side is taken 5% in excess and the other 6% in deficit. Find the error percent in area calculated, of the plot.

Answer.  1.3%.

Detailed Solution. Let the length and breadth of the rectangle be l and b units respectively.

Then, correct area = (lb) sq. units.

Calculated area = ((105 l × 100) / (94 l × 100)) sq. units. = (987 lb / 1000) sq. units.

Error in measurement = (lb - (987/1000).lb) sq. units = (13lb/1000) sq. units.

∴ Error% = (13lb/1000 × 1/lb × 100)%a = 1.3%.

Q9. Instead of walking along two adjacent sides of a rectangular field, a boy took a shortcut along the diagonal of the field and saved a distance equal to half of the longer side. Find the ratio of the shorter side of the rectangle to the longer side

Answer.  3 : 4.

Detailed Solution. Let the length of the longer side of the field be l and that of the shorter side be b.

Then, diagonal = √l² + b².

∴ (l + b) - √l² + b² = (1/2).l → √l² + b² = 1/2 + b

→ 2√l² + b² = l + 2b → 4(l² + b²) = l² + 4b² + 4lb

→ 3l² = 4lb → 3l = 4b → ^b/_l = 3/4

Hence, required ratio = 3 : 4.

Q10. A rectangular grassy plot 110 m by 65 m has a gravel path 2.5 m wide all round it on the inside. Find the cost of gravelling the path at 80 paise per sq. metre.

Answer.  Cost of gravelling the path =   ₹ (850 × 80/100) × ₹ 680.

Detailed Solution. Area of the plot = (110 × 65) m² = 7150 m²

Area of the plot excluding the path = [(110 - 5) × (65 - 5)] m² = 6300 m²

∴  Area of the path = (7150 - 6300) m² = 850 m²

Cost of gravelling the path =   ₹ (850 × 80/100) × ₹ 680.

Q11. The diagonal of a rectangular field is 15 m and its area is 108 sq. m. What will be the total expenditure in fencing the field at the rate of ` 5 per metre?

Answer. ₹ 210.

Detailed Solution. Let the length and breadth of the rectangle be l and b metres respectively.

Then, √l² + b² = 15 and lb = 108 ⇒ l² + b² = 225 and lb = 108

⇒ (l + b)² = (l² + b²) + 2lb = 225 + 216 = 441

⇒ l + b = √441 = 21.

∴  Perimeter of the field = 2 (l + b) = (2 × 21) m = 42 m.

Hence, cost of fencing = ₹ (42 × 5) = ₹ 210.

Q12. The perimeters of two squares are 40 cm and 32 cm. Find the perimeter of a third square whose area is equal to the difference of the areas of the two squares.

Answer. 24 cm.

Detailed Solution. Side of first square = (40/4) cm ≈ 10 cm; Side of second square = (32/4) cm ≈ 8 cm.

Area of third square = [(10)² - (8)² ] cm² = (100 - 64) cm² = 36 cm²

Side of third square = √36 cm ≈ 6 cm.

∴ Required perimeter = (6 × 4) cm = 24 cm.

Q13. The length of a rectangle R is 10% more than the side of a square S. The width of the rectangle R is 10% less than the side of the square S. What is the ratio of the area of R to that of S?

Answer. 99:100.

Detailed Solution. Let each side of the square S be x units.

Then, length of rectangle R = 110% of x = (11x/10) units.

And, width of rectangle R = 90% of x = (9x/10) units.

∴ Ratio of areas of R and S = ((11x × 10x) x (9x/10)) : x² = (99x²/100) : x² = 99:100.

Q14. Find the largest size of a bamboo that can be placed in a square of area 100 sq. m.

Answer. 14.14 m.

Detailed Solution. Side of the square = √100 m = 10 m.

Largest size of bamboo = Length of diagonal of the square

= 10 √2 m. = (10 × 1.414) m = 14.14 m.

Q15. A rectangular courtyard, 3.78 m long and 5.25 m broad, is to be paved exactly with square tiles, all of the same size. Find the least number of square tiles covered.

Answer. 450.

Detailed Solution. Area of the room = (378 × 525) cm².

Size of largest square tile = H.C.F. of 378 cm and 525 cm = 21 cm.

Area of 1 tile = (21 × 21) cm².

∴ Number of tiles required = ((378 × 525) / (21 × 21)) ≈ 450.

Q16. Find the area of a square, one of whose diagonals is 3.8 m long.

Answer. 7.22 m²

Detailed Solution. Area of the square = ½ × (diagonal)² = (½ × 3.8 × 3.8) m² = 7.22 m².

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