# 1000+ Goldman Sachs Aptitude Questions and Answers Pdf - 1

Question: 1

A rectangular park 60 m long and 40 m wide has two concrete crossroads running in the middle of the part and rest of the park has been used as a lawn. If the area of the lawn is 2109 sq.m, then what is the width of the road?

(A) 2 m

(B) 2.91 m

(C) 3 m

(D) 5.82 m

Ans; C

Area of the park = (60 × 40) m^{2} = 2400 m^{2}

Area of the lawn = 2109 m^{2}

∴ Area of the crossroads = (2400 - 2109) m^{2} = 291 m^{2}

Let the width of the road be x metres. Then

60x + 40x – x^{2} = 291

⇔ x^{2} - 100x + 291 = 0 ⇔ (x = 97)

⇔ (x – 3) = 0 ⇔ x = 3.

Question: 2

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.

(A) 10 cm

(B) 12 cm

(C) 24 cm

(D) 36 cm

Ans: C

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}]cm^{2} = (100 - 64) cm^{2} = 36 cm^{2}.

Side of third square = $$√36$$ cm = 6 cm.

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

Question: 3

The diameter of the driving wheel of a bus is 140 cm. How many revolutions per minute must the wheel make in order to keep a speed of 66 kmph?

(A) 100

(B) 150

(C) 250

(D) 350

Ans: C

Distance to be covered in 1 min = $$({66 × 1000} / {60})$$m = 1100 m.

Circumference of the wheel = $$(2 × {22}/{7} × 0.70)$$m = 4.4 m.

∴ Number of revolutions per min. $$({1100} / {4.4})$$ = 250.

Question: 4

The sides of a triangle are in the ratio of $${1/2} : {1/3} : {1/4}$$. If the perimeter is 52 cm, then the length of the smallest side is

(A) 9 cm

(B) 12 cm

(C) 14 cm

(D) 16 cm

Ans: B

Ratio of sides = $${1/2} : {1/3} : {1/4}$$ = 6 : 4 : 3.

Perimeter = 52 cm. So, sides are

$$(52 × {6}/{13})$$cm, $$(52 × {4}/{13})$$ cm and $$(52 × {3}/{13})$$ cm.

So, a = 24 cm, b = 16 cm, c = 12 cm.

∴ Length of smallest side = 12 cm.

Question: 5

The circumference of a circular race track, 14 m wide, is 440 m. Find the radius of the outer circle.

(A) 64

(B) 74

(C) 84

(D) 94

Ans: C

Let inner radius be r metres.

Then, 2Πr = 440 ⇒ r = $$(440 × {7}/{44})$$ = 70 m.

∴ Radius of outer circle = (70 + 14) m = 84 m.

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