GRE
Practice Test 39
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Spider is trying to climb a wall 30 inches high, it climbs 10 inches in 5 minutes and then rest for 5 minutes during which it slips back 5 inches.
Time taken by the spider to reach the top minutes 60
  • Solution

    Spider climbs 10 inches in 5 minutes and slips 5 inches in next 5 minutes. Thus, he climbs net 5 inches in 10 minutes. In this manner he will climb 20 inches in 40 minutes.

    Thereafter in next 5 minutes he will climb 10 inches and will reach the top. Thus he takes 40 + 5 = 45 minutes to reach the top.

    ∴ Col. A = 45

Sum of the first 43 terms of the series
1⁄3 + 1⁄4 – 1⁄5 – 1⁄3 – 1⁄4 + 1⁄5 + 1⁄3 + 1⁄4 – 1⁄5….. \(\frac{7}{12}\)
  • Solution

    One can observe that fractions get repeated after certain interval.

    Col. A:

    The first six terms of the series add to (1⁄3 +1⁄4-1⁄5) +(-1⁄3- 1⁄4+1⁄5)= 0

    ∴ The sum of first 42 terms is zero as 7 such patterns are completed.

    43rd term = 1⁄3

    ∴ Col.A = 1⁄3 = \(\frac{4}{12}\),Col.B = \(\frac{7}{12}\)

 

Unit’s digit in the sum (344)153 + (344)154 Unit’s digit in the no. (345)152
  • Solution

    Consider the power cycle for number 4

    41 = 1,42 = 1\(\underline{6}\),43 = 61,44 = 25\(\underline{6}\)

    ∴ If the index is odd, number ends in 4.

    If the index is even, number ends in 6 .

    ∴ (344)153 ends in 4 and (344)154 ends in 6

    ∴ Col A = 0

    Col. B = Any power of 5 ends in 5. ∴ Col.B = 5

Fresh grapes contain 75% water while dry grapes contain 20% water. Weight of fresh grapes is 640 kgs.
Total weight in kgs when fresh grapes are converted in dry grapes 150
  • Solution

    Pulp in fresh grapes of 640 kgs will be 25% i.e 160 kgs. Dry grapes contain 20% water and 80% pulp. ∴ In the pulp of 180 kgs, 40 kgs water will be added and the will get 200 kgs of dry grapes

 

\(\sqrt[3]{6\sqrt{2}}\) \(\sqrt[3]{6\sqrt{2}}\)
  • Solution

    √2 = 1.414 ∴ 6√2 = 8.5 Aprox.

    ColA:\(\sqrt[3]{8.5}\)

    ColA :\(\sqrt[3]{9}\)

To travel 600 Ian, Bob takes 8 hours more than Abel. If however, speed of Bob is doubled, he takes 2 hours less than Abel.
Abel’s speed in kmph 60
  • Solution

    Let Bob's and Abel's speed be x and y respectively.

    ∴ \(\frac{600}{x}-\frac{600}{y}\)=8.....(I)

    &\(\frac{600}{2x}-\frac{600}{y}\)= -2.....(II)

    Subtracting (II)from (I)

    \(\frac{600}{x}-\frac{600}{2x}\)= 10

    ∴ \(\frac{600}{2x}\) =10

    ∴ x = 30 kmph

    ∴ y = 50kmph

    ∴ Col. A = 50, Col. B = 60

 

\(\sqrt{\frac{3}{25}}+\sqrt{\frac{3}{25}}+\sqrt{\frac{3}{25}}\) \(\sqrt{\frac{3}{9}}+\sqrt{\frac{3}{9}}+\sqrt{\frac{3}{9}}\)
  • Solution

    Equating the base

    Col.A = \(\sqrt{\frac{27}{225}}+\sqrt{\frac{27}{225}}+\sqrt{\frac{27}{225}}\)

    Col.B = \(\sqrt{\frac{50}{225}}+\sqrt{\frac{50}{225}}+\sqrt{\frac{50}{225}}\)

 

The volumes of s here with radius 6 cm. The volume of cube with side 12 cm
  • Solution

    Col.A = 4⁄3π(6)3 = 4 × π × 36 × 2

    = 12 × 12 × 2π = 12 × 12 × 2 × 3.14

    Col.B = (12)3 = 12 × 12 × 12

 

The number of primes of which 7 is an integral multiple The number of primes of which 17 is an integral multiple
  • Solution

    Col. A : 7 is a prime number.

    ∴ It has exactly 2 factors, viz. number 1 and number 7 itself.

    ∴ 7 is the only factor and prime factor of 7.

    Col. A = 1

    Col. B = 1.....by the same theory

In the figure above, 3 circles are tangent to each other at points shown. Circle P has a diameter 10 cm.,circle Q has 8 cm. and circle R has 6 em.

Perimeter of ΔPQR in cm Area of ΔPQR in cm2
  • Solution

    Radius of circles with centre P, Q, R is 5 cm., 4 cm. and 3 cm. respectively.

    ∴ PQ = 5 + 4 = 9 cm.

    PR = 5 + 3 = 8 cm.

    QR = 4 + 3 = 7 cm.

    ∴ Col. A = 9 + 8 + 7 = 24 cm.

    Col. B = Semi perimeter(s) of ΔPQR = 12 cm.

    By Hero's formula

    Area = \(\sqrt{s(s - a)(s - b)(s - c)}\)

    = \(\sqrt{12 \times (12- 9)(12 - 8)(12 -7)}\)

    = \(\sqrt{12\times 12\times 5}=12\sqrt{5}\)

    Col.B = 12√5

    Col.A = 24 = 12 × 2 = 12√4

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