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Quantitative Ability Test - 4

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Quantitative Ability Test - 4
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Weekly Quiz Competition
  • Question 1
    4 / -1

    There are 3050 employees in an organisation. Out of which 48% got transferred to different places and 1159 new employees joined the organization in the new year. The total number of employees in the organization now is what percentage of the original number of employees?

    Solution
    Number of employees transferred \(=48 \%\) of 3050 \(=1464\)
    Total number of employees in the new year \(=(3050-1464)+1159=2745\)
    \(\therefore\) Required percentage \(=\frac{2745}{3050} \times 100-90 \%\)
  • Question 2
    4 / -1

    A mixture of 729 ml contains milk and water in the ratio 7 : 2. How much more water should be added to get a new mixture containing milk and water in the ratio 7 : 3?

    Solution
    The amount of milk \(=\frac{7}{9} \times 729=567 \mathrm{m}\) and the amount of water will be \(=(729-567)=162 \mathrm{ml}\) Let \(x^{*}\) m of water is added to the moture.
    So, \(\frac{567}{162+x}=\frac{7}{3} \Rightarrow x=81 \mathrm{ml}\)
  • Question 3
    4 / -1

    Aman and Chintu can complete a job in 40 days and 25 days respectively. How long will they take to complete the work if they work on alternate days?

    Solution

    We cannot determine the required number of days as it is not known that who among the two started the work.

  • Question 4
    4 / -1

    A sum was borrowed at 20% p.a. compound interest. It was repaid in 3 annual installments with each installment being paid at the end of a year. The first, second and third installments were Rs. 1200, Rs. 1152 and Rs. 2592 respectively. Determine the sum borrowed.

    Solution

    The value of the first installment (at the time the sum was borrowed) = 1200 / 1.2 = 1000

    The value of the second installment (at the time the sum was borrowed) = 1152 / (1.2)2= 800

    The value of the third installment (at the time the sum was borrowed) = 2592 / (1.2)3= 1500

    Sum borrowed = Total value (at the time the sum was borrowed) = Rs. 3300.

  • Question 5
    4 / -1

    The average age of a man and his son is 40 years. The ratio of their ages 15 years hence is 7 : 4 respectively. If the ratio of the present ages of the man and his wife is 11 : 9. What is the age of his wife?

    Solution
    Let the ages of the man and his son 15 years hence be \(7 x\) and \(4 x\) years respectively. According to the question, \(7 x+4 x=2 \times 40+15+15\)
    \(\Rightarrow 10 x=110\)
    \(\Rightarrow x=10\)
    \(\therefore\) Man's present age \(=7 x-15=7 \times 10-15=55\) years. Let the present ages of the man and his wife be \(11 x\) and \(9 x\) years respectively. Now \(11 x=55\) years Age of his wite \(=45\) years.
  • Question 6
    4 / -1

    The sum of four numbers is 64. If you add 3 to the first number, 3 is subtracted from the second number, the third is multipled by 3 and the fourth is divided by 3, then all the results are equal. What is the difference between the largest and the smallest of the original numbers?

    Solution
    Solution:
    Let the four numbers be \(x, y, z\) and \(w\) Given, \(x+y+z+w=64\)
    and \(x+3=y-3=3 z=\frac{w}{3}\)
    \(\therefore y=x+6 ; z=\frac{x+3}{3} ; w=3(x+3)\)
    Substruting the values \(: x+x+6+\frac{x+3}{3}+3(x+3)\)
    \(=64\)
    \(\Rightarrow 5 x+\frac{x+3}{3}=49\)
    Solving this, \(\underline{x}=9\) \(y=x+6=15\)
    \(z=\frac{x+3}{3}=4\)
    \(w=3(x+3)=36\)
    Required difference \(=36-4=32\)
  • Question 7
    4 / -1

    The ratio of the present ages of a mother and daughter is 7:1. Four years ago the ratio of their ages was 19:1. What will be the mother's age four years from now?

    Solution
    Let present ages of mother and daughter be \(7 x\) and \(x\)
    Then \(, \frac{7 x-4}{x-4}=\frac{19}{1}\)
    \(\Rightarrow 7 x-4=19 x-76\)
    \(\Rightarrow 12 x=72\)
    \(\Rightarrow x=6\)
    \(\therefore\) Mother's age 4 years hence \(=7 \times 6+4=46\) years.
  • Question 8
    4 / -1

    Which of the following is a factor of 5! + 6! + 7! + 8! + 9!?

    Solution

    5! + 6! + 7! + 8! + 9! = 5!(1 + 6 + 6 x 7 + 6 x 7 x 8 + 6 x 7 x 8 x 9)

    = 5!(1 + 6 + 42 + 336 + 3024)

    = 5! x 3409

    = 120 x 3409

    = 120 x 7 x 487

    Therefore, 120 x 487 = 58440 is a factor of the number.

    Hence, (A).

  • Question 9
    4 / -1

    In triangle PQR, PQ = 12 units and PR = 16 units. Point S is on side QR such that Q-S-R and PQ = PS. M is the midpoint of side PR. What is the distance between the midpoints of segments QS and PR?

    Solution

    Suppose T is the midpoint of QS.

    PS = PQ or Δ PQS is an isosceles triangle. Therefore, PT is perpendicular to QS. Consider Δ PTR. Angle T is 90 degrees. Therefore, T lies on the circumference of the circle with PR as diameter or diameter of the circle is 16. MT is the radius of the circle. Therefore, the distance between the midpoints of segments QS and PR is equal to radius of the circle = 8 units. Hence, (C).

  • Question 10
    4 / -1

    The angle subtended by an arc at the centre of a circle is 40° If the area of the sector formed by the arc and the radii of the circle is 68 4/9 sq. cms, find the radius of the circle (take π=22 /7)

    Solution
    Let the radius of the circle be r \(\mathrm{cm}\).
    \[
    \frac{40^{\circ}}{300^{-5}}\left(\frac{22}{7} r^{2}\right)=\frac{616}{9}
    \]
    \(\mathrm{r}=14 \mathrm{cm}\)
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