(Paper) GMAT Math Problem Solving & DS Practice Questions Paper

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Algebra

1. A poultry farm has only chickens and pigs. When the manager of the poultry counted the heads of the stock in the farm, the number totaled up to 200. However, when the number of legs was counted, the number totaled up to 540. How many chickens were there in the farm?

  1. 70
  2. 120
  3. 60
  4. 130
  5. 80

The correct choice is (4) and the correct answer is 130.

Explanatory Answer:
Let there by 'x' chickens and 'y' pigs.
Therefore, x + y = 200 --- (1)
Each chicken has 2 legs and each pig has 4 legs
Therefore, 2x + 4y = 540 --- (2)
Solving equations (1) and (2), we get x = 130 and y = 70.
There were 130 chickens and 70 pigs in the farm.
 

2. Three years back, a father was 24 years older than his son. At present the father is 5 times as old as the son. How old will the son be three years from now?

  1. 12 years
  2. 6 years
  3. 3 years
  4. 9 years
  5. 27 years

The correct choice is (4) and the correct answer is 9 years.

Explanatory Answer:
Let the age of the son 3 years back be x years
Therefore, the age of the father 3 years back was x + 24
At present the age of the son is x + 3 and the father is 5 times as old as the son.
i.e., x + 24 + 3 = 5(x + 3)
i.e., x + 27 = 5x + 15
or 4x = 12 or x = 3.

Therefore, the son was 3 years old 3 years back and he will be 9 years old three years from now.

3. For what values of 'k' will the pair of equations 3x + 4y = 12 and kx + 12y = 30 not have a unique solution?

  1. 12
  2. 9
  3. 3
  4. 7.5
  5. 2.5

The correct choice is (2) and the correct answer is 9.

Explanatory Answer:
A system of linear equations ax + by + c = 0 and dx + ey + g = 0 will have a unique solution if the two lines represented by the equations ax + by + c = 0 and dx + ey + g = 0 intersect at a point.

That is, if they are not parallel lines. i.e., the two lines should have different slopes.
ax + by + c = 0 and dx + ey + g = 0 will not represent two parallel lines if their slopes are different.
i.e., when a/d not equal to b/e
In the question given above, a = 3, b = 4, d = k and e = 12.
Therefore, k not equal to 9 or 'k' should not be equal 9 for the pair of equations to have a unique solution.

In other words, when k = 9, the system of equation will not have any solution as the two lines represented by the equations will be parallel lines.

4. The basic one-way air fare for a child aged between 3 and 10 years costs half the regular fare for an adult plus a reservation charge that is the same on the child's ticket as on the adult's ticket. One reserved ticket for an adult costs $216 and the cost of a reserved ticket for an adult and a child (aged between 3 and 10) costs $327. What is the basic fare for the journey for an adult?

  1. $111
  2. $52.5
  3. $210
  4. $58.5
  5. $6

The correct choice is (3) and the correct answer is $210.

Explanatory Answer:
Let the basic fare for the child be $X.
Therefore, the basic fare for an adult = $2X.
Let the reservation charge per ticket be $Y
Hence, an adult ticket will cost 2X + Y = $216
And ticket for an adult and a childe will cost 2X + Y + X + Y = 3X + 2Y = 327
Solving for X, we get X = 105.

The basic fare of an adult ticket = 2X = 2*105 = $210

Arithmetic Progression and Geometric Progression

1. What is the sum of all 3 digit numbers that leave a remainder of '2' when divided by 3?
  1. 897
  2. 164,850
  3. 164,749
  4. 149,700
  5. 156,720
The correct choice is (B) and the correct answer is 164,850.

 
Explanatory Answer:
The smallest 3 digit number that will leave a remainder of 2 when divided by 3 is 101.
The next number that will leave a remainder of 2 when divided by 3 is 104, 107, ....
The largest 3 digit number that will leave a remainder of 2 when divided by 3 is 998.
So, the given series is an AP with the first term being 101 and the last term being 998 and thhe common difference being 3.
Sum of an AP = ap summation formula

We know that in an A.P., the nth term an = a1 + (n - 1)*d
In this case, therefore, 998 = 101 + (n - 1)* 3
i.e., 897 = (n - 1) * 3
Therefore, n - 1 = 299
Or n = 300.
Sum of the AP will therefore, be 101 + 998 /2 * 300 = 164,850

2. How many 3 digit positive integers exist that when divided by 7 leave a remainder of 5?

  1. 128
  2. 142
  3. 143
  4. 141
  5. 129
The correct choice is (E) and the correct answer is 129.

Explanatory Answer:
The smallest 3-digit positive integer that when divided by 7 leaves a remainder of 5 is 103.
The largest 3-digit positive integer that when divided by 7 leaves a remainder of 5 is 999.
The series of numbers that satisfy the condition that the number should leave a remainder of 5 when divided by 7 is an A.P (arithmetic progression) with the first term being 103 and the last term being 999 having a common difference of 7.
We know that in an A.P, 'l' the last term is given by l = a + (n - 1) * d, where 'a' is the first term, 'n' is the number of terms of the series and 'd' is the common difference.
Therefore, 999 = 103 + (n - 1) * 7
Or 999 - 103 = (n - 1) * 7
Or 896 = (n - 1) * 7
Or n - 1 = 128
Or n = 129

3. The average of 5 consecutive integers starting with m as the first integer is n. What is the average of 9 consecutive integers that start with m+2?
  1. m + 4
  2. n + 6
  3. n + 3
  4. m + 5
  5. n + 4
The correct choice is (E) and the correct answer is n + 4.

Explanatory Answer:
The fastest way to solve problems of this kind is to take numerical examples.
The average of 5 consecutive integers from 1 to 5 is 3. Therefore, the value of m is 1 and the value of n is 3.
Now, the average of 9 consecutive integers starting from m + 2 will be average of integers from 3 to 11.
The average of numbers from 3 to 11 is 7.
Now look at the answer choices. Only choice (E) satisfies this condition.

4. The sum of the fourth and twelfth term of an arithmetic progression is 20. What is the sum of the first 15 terms of the arithmetic progression?

  1. 300
  2. 120
  3. 150
  4. 170
  5. 270
The correct choice is (C) and the correct answer is 150.

Explanatory Answer:
The sum of the 4th and 12th term = 20.
Let t1 be the first term, t4 be the fourth term, and t12 be the 12th term
Then t4 + t12 = 20
t4 can be expressed as t1 + 3d
Similarly, t12 can be expressed as t1 + 11d
Then t4 + t12 = 20 can be expressed as   t1 + 3d + t1 + 11d = 20
=>      2t1 + 14d = 20
=>      t1 + 7d =10
=>      t8 = 10
The sum of the first 15 terms = 15/2(t1 + t15)
In an arithmetic progression t1 + t15 = t1 + t1 + 14d = 2t1 + 14d = 2(t1 + 7d) = 2(t8).
Therefore, the sum of the first 15 terms =15/2(2*10) = 150

5. If the ratio of the sum of the first 6 terms of a G.P. to the sum of the first 3 terms of the G.P. is 9, what is the common ratio of the G.P?
  1. 3
  2. 1/3
  3. 2
  4. 9
  5. 1/9
The correct choice is (C) and the correct answer is 2.

Explanatory Answer:
The sum of the first n terms of a G.P. is given by , where 'a' is the first term of the G.P., 'r' is the common ratio and 'n' is the number of terms in the G.P.
Therefore, the sum of the first 6 terms of the G.P will be equal to
And sum of the first 3 terms of the G.P. will be equal to
The ratio of the sum of the first 6 terms : sum of first 3 terms = 9 : 1
i.e.
=>
=> r3 + 1 = 9
=> r3 = 8
=> r = 2

6. Set A contains all the even numbers between 2 and 50 inclusive. Set B contains all the even numbers between 102 and 150 inclusive. What is the difference between the sum of elements of set B and that of set A?

  1. 2500
  2. 5050
  3. 11325
  4. 6275
  5. 2550
The correct choice is (A) and the correct answer is 2500.

Explanatory Answer:
SET 1 - (2, 4, 6, 8,...., 50). A total of 25 consecutive even numbers.
SET 2 - (102, 104, 106,....., 150). Another set of 25 consecutive even numbers.
Difference between 1st term of set 1 and set 2 is 100. Difference between 2nd term of set 1 and set 2 is 100 and so on.
So total difference is (100 + 100 + 100 + ....) = 25*100 = 2500.

More Question:

Inequalities
Inequalities is considered to be one of the hard topics in the problem solving section. Questions involving algebraic expression, modulus, exponents in inequalities are covered.

Geometry
Triangles, similar triangles, right triangles, equilateral triangles, lines, parallel lines, angles, quadrilaterals, circles.

Mensuration - Solid Geometry
Area, volume, surface area, of 3d objects such as cuboids, cubes, spheres, hemispheres, cones, cylinders, pyramids and prisms.

Coordinate Geometry
Slopes, intercepts, quadrants, distance between points, midpoint theorem, coordinates of Centroid, incenter. Finding area of a triangle given coordinates of its vertices.

Number Properties, Theory & Number Systems
LCM, HCF, remainders, factorials, unit digits of numbers, word problems.

Percentages
Percentages to fraction or decimal conversion and the reverse. Problems using the concept of percentages.

Counting Methods (Permutation, Combination) and Discrete Probability
Permutation, combination, sampling with replacement, ordering, re-arrangement of letters of a word, seating arrangements.

Profits
Profit, loss, expressed as a percentage of cost price, discounts, marked price.

Quadratic Equations
Nature of roots, real roots, imaginary roots, equal roots, solution to quadratic equations, word problems that require framing quadratic equutions to get the solution.

Ratio, Proportion and Variance
Standard problems that involve ratios and proportion, continued proportions, direct and inverse variations.

Sets (Set Theory)
Union, intersection of sets, disjoint sets, mutually exclusive sets.

Simple and Compound Interest
Simple interest, compound interest, compounding periods other than on an annual basis.

Rates : Speed, Time and Distance
Average speed, relative speed, train problems, boats in streams, races.

Rates : Work - Time and Pipes - Cisterns
Amount of work done by a man or woman, time taken by pipes to fill a cistern.

Data Sufficiency
About a third of the questions in the GMAT quant section are data sufficiency questions.