 # 15 SAT Maths Practice Questions

## Question 1 If O is the centre of the circle shown above and OP = PQ = QR, what is the value of x?

a) 15

b) 30

c) 45

d) 60

We are given that OP = PQ = QR.

And, OP = OR (radii of the circle).

So, OP = PQ = QR = OR

So, all the sides of the quadrilateral OPQR are equal. Hence, it will be a rhombus.

Note: We cannot assume OPQR to be a square, because nothing is mentioned about the angles. So, we can’t assume angle OPQ to be 90°, hence we can’t assume x to be 45°.

if all the sides are equal in a quadrilateral then all we can surely conclude is that the Quadrilateral will be a Rhombus. Also, the diagonals of a rhombus are angle bisectors.

Now, if we join OQ which is also the radius of the circle, then OP = OQ = PQ. Therefore, triangle OPQ will be an equilateral triangle. Hence, angle OPQ = 60°.

So, x = (angle OPQ) / 2 = 30°. (Diagonals of a rhombus are angle bisectors)

TIP: We cannot assume anything by sight or by measurement unless or until it is not given in the question.

OR

METHOD 2:

Angle POQ = Angle ROQ = 60°. (As both the triangles OPQ and OQR are equilateral triangles).

So, Angle POR = 120°.

In triangle OPR;,/p> OP = OR (radii of the circle)

So, angle OPR = angle ORP = x (In a triangle, angles opposite to equal sides are equal)

Angle POR + Angle OPR + Angle ORP = 180 (Sum of internal angles of a triangle is 180°)

120 + x + x = 180

x = 30

ANS (b) = 30

## Question 2 If two sides of a triangle are 6 and 8, which of the following cannot be the area of the triangle?

a) 0.01

b) 5

c) 10

d) 25 Now, we have to find which of the following cannot be the area of this triangle.

Let’s consider the figure above, AB = 6 units & BC = 8 units.

As we do not know the height, we will not be able to calculate the exact area.

However, in such questions when we need to select which of the options can / cannot be the answer, you would find that eliminating the options is one of the convenient approaches. we know that Area of Triangle = (1/2) × base × height   So, in the above figure as the base is fixed i.e. 8 so area will only be maximum when the height will be maximum

Or we could state that the area will be maximum if the other side 6 becomes the height.

Hence, Maximum Area = (1/2) × 8 × 6/

= 24.

As any option more than 24 cannot be the possible value of area. Hence 25 cannot be the area of the triangle with dimensions of 6 and 8 units.

Also note that the area will be minimum when the height will be minimum.

Height will be minimum if both the sides AB & AC coincides with the base BC, then area will become 0.

So, it can take any value above 0 to form a triangle.

0 < Area of Triangle ≤ 24

Hence, area of triangle cannot be 25.

ANS (D)

### Question 3: Which of the following can be the graph of the line L, ax + by + a = 0, if a b > 0?

a) b) c) d) The standard form of equation of a line is: y = m x + c

Where,

m = slope of the line

c = y – intercept (the y-coordinate of the point where the line intersects the y-axis)

So, the equation ax + by + a = 0 can be re-written in the form of y = m x + c as:

y = ((-a)/b) x + ((-a)/b)

So, m = ((-a)/b)

c = ((-a)/b)

We are given that, (a b) > 0

So, (a/b) > 0

(-a)/b < 0

Hence,

1) m < 0 → slope is negative; which means it will be a descending line (a line which descends when we move from negative x-axis to positive x-axis).

2) C < 0 → y- intercept is negative; which means line will intersect the y-axis below the origin.

So, this will be a descending line intersecting the y-axis below the origin.

## Question 4: If a and b are positive integers, and a + b < 15, then what is the greatest possible value of b – a?

ANS:

>a and b are positive integers.

>a + b < 15

We have to find the greatest possible values of (b – a).

Now, (b – a) will only be greatest when b will take its maximum value and a will take its minimum value.

As, a and b are defined as positive integers, so (a)min = 1

By, putting the value of a in: a + b < 15

1 + b < 15

b < 14

But we are looking for the maximum value of b.

So, maximum value of positive integer b < 14 will be 13.

(b)max = 13

The greatest value of (b – a) = (b)max - (a)min

= 13 -1

= 12

ANS = 12

## Question 5

| b - 8 | = 12

|b - 13 | = 17

If b satisfies both the equations above. What is the value of |b|?

GIVEN:

|b – 8| = 12

|b – 13| = 17

CONCEPT: •

• |x| = |x – 0|; signifies the distance of ‘x’ from the origin on the number line.

• |x – a|; signifies the distance of ‘x’ from another point ‘a’ on the number line.

So,

1) |b – 8| = 12

The above equation signifies that, we are looking for those values of b whose distance from 8 is equal to 12 units on the number line. From the number line drawn above, the possible values of b will be (-4) and 20.

2) |b – 13| = 17

The above equation signifies that, we are looking for those values of b whose distance from 13 should be equals to 17 units on the number line. From the number line drawn above, the possible values of b will be (-4) and 30.

Now, as b is satisfying both the equations above, so the only possible value of b = - 4

But we have to find the value of |b|;

|b| = |-4| = 4

ANS = 4

## Question 6: Train A’s average speed for its journey is at least 70 miles/hour and at most 100 miles/hour. If the journey is 7000 miles long, then which of the following equations represents all possible values of the time taken, t, by the train to complete the journey?

a) |t – 15|≥ 85

b) |t – 15|≤ 85

c) |t – 85|≤ 15

d) |t – 85|≥ 15

We know, Time = (Distance )/Speed

The minimum possible amount of time taken by the train would be when train travels the entire distance at the maximum possible speed, i.e. 100 mph.

t minimum = 7000/100 = 70 hours.

The maximum time taken by the train would be when it travels the entire distance with the minimum possible speed.

tmaximum = 7000/70 = 100 hours.

In other words, it can be said that time taken by the train, t will satisfy the following inequality:

70 ≤ t ≤ 100

One way to go further would be solving each inequality in the options and finding out which of the 4 options gives the above mentioned result for t.

However, we recommend you follow the following step by step reasoning-based approach.

We know, 70 ≤ t ≤ 100

STEP 1: Take the average of the two extremities i.e. 70 & 100, which comes out to be 85.

STEP 2: This average, 85, becomes the reference point for the distance form. In other words, L.H.S of the inequality would be |t – 85|

STEP 3: Calculate the distance between the extremities and the average.

|100 – 85| = 15 & |70 – 85| = 15

This becomes the R.H.S of the inequality

|t – 85| ? 15

STEP 4: To find what fills the question mark above, just observe that the range for t, 70 ≤ t ≤ 100

L.H.S ≤ R.H.S

So, the inequality will be:

|t – 85| ≤ 15

ANS: C

## Question 7:

Today Tina is thrice as old as Rita and Rita is 4 years younger than Agatha. If Tina, Rita, and Agatha are all alive 5 years from today, which of the following must be true on that day?

1. Tina is thrice as old as Rita.
2. Rita is 4 years younger than Agatha.
3. Agatha is older than Tina.

a) II only

b) III only

c) II and III only

d) I, II and III

In this question, Tina is thrice as old as Rita and Rita is 4 years younger than Agatha.

TIP: Always try to introduce as few variables as possible.

So, Let the Present age of Agatha be ‘A’.

Then, the Present age of Rita will be = A - 4.

And the Present age of Tina will be = 3 (A – 4) = 3A - 12.

 TINA RITA AGATHA PRESENT AGE 3A - 12 A - 4 A AGE AFTER 5 YEARS (3A – 12) + 5 = 3A - 7 (A – 4) + 5 = A + 1 A + 5

Now we have to find which of the following must be true on the day, 5 years from today.

1) Tina is thrice as old as Rita.

The age of Tina = 3A – 7

And, the age of Rita = A + 1

And, as 3A – 7 ≠ 3(A + 1)

So, Tina is not thrice as old as Rita.

Hence, this statement is not true.

2) Rita is 4 years younger than Agatha.

Age of Rita = A + 1

Age of Agatha = A + 5

So, Rita is 4 years younger than Agatha.

Hence, this statement must be true.

3) Agatha is older than Tina.

Age of Agatha = A + 5

Age of Tina = 3 A – 7

Now for Agatha to be older than Tina

A + 5 > 3A – 7

12 > 2A

A < 6

Now we are not given any information of Agatha’s Age A with regards to 6 years. So, the statement “Agatha is older than Tina” is not always true. So, this statement is not must be true.

ANS: (a) II only.

## Question 8: If the number n of calculators sold per week varies with the price p in dollars according to the equation n = 300 – 20p, what does the constant ‘20’ signify in the equation?

a) The change in the revenue with each \$1 increase in the price.

b) The change in the number of calculators sold with the increase of \$1 in the price.

c) The increase in the revenue with each \$1 increase in the price.

d) The increase in the number of calculators sold with the increase of \$1 in the price.

The standard form of equation of a line is: y = m x + c

Where,

m = slope of the line = (Change in Y-coordinate)/(Change in X-coordinate)

c = y – intercept (the y-coordinate of the point where the line intersects the y-axis)

Basically, c is the value of y when we put x = 0 in the equation.

Now, the given equation is: n = 300 – 20p

Or, n = -20p + 300

So, if we compare this with the equation of a line, then;

y = n; number of calculators

x = p; price of each calculator

m = -20; which represents the slope; which basically signifies the change in number of calculators for a unit change in price.

c = 300; which is the y –intercept.

We have to explain what 20 is signifying in the equation.

As, we have mentioned above 20 is the slope so it is signifying the change in number of calculators sold for an increase of \$1 in the price.

So, Answer should be option (b)

Sometimes by mistake, we mark the answer as option (d) but if we analyze the value of m = -20 which means the slope is negative which in turn signifies that there will be a decrease in the number of calculators sold for an increase of \$1 in the price.

## Question 9: If f(x) = (x – 3) (x + 5), which of the following is an equivalent form of the function f, in which the minimum value of f appears as a constant or a coefficient?

a) f(x) = x2 – 15

b) f(x) = x2 – 2x – 15

c) f(x) = (x – 1)2 – 14

d) f(x) = (x + 1)2 – 16

f(x) = (x – 3) (x + 5) = x2 + 2x - 15

We have to find out of the given options which one is an equivalent form of the function f, in which the minimum value of f appears as a constant or coefficient.

Now, we know that f(x) is a quadratic function, so graphically this will be a parabola.

As, Coefficient of x2 is 1 which is a positive value, hence it will be an upward parabola

And, In the case of an upward parabola, the function will take its minimum value on the vertex.

So, the y-coordinate of the vertex will represent the minimum value of the function.

Now, the equivalent form in which the coordinates of vertex can appear as a constant or coefficient is given as:

f (x) = a (x – h)2 + k

where, (h, k) are the coordinates of the vertex. Now, f(x) = (x – 3) (x + 5)

= x2 +2x - 15

= x2 +2x + 12 – 12 - 15 [adding and subtracting 12 to complete the whole square; in the form of (x – h)2]

= (x +1)2 – 16

Hence, the answer will be option D.

ANS (D)

## Question 10: If f(x) = x2 + b x + 2b2, which of the following could be the graph of y = f(x)?

a) b) c) d) CONCEPT:

In a general quadratic function f(X) = AX2 + B X + C

> A :-

If A > 0 => Parabola will be upward in shape.

And, If A < 0 => Parabola will be downward in shape.

> C:-

C signifies the y- intercept of the parabola which is the y – coordinate of the point where the parabola will intersect the y – axis.

If C > 0 => Parabola will intersect the y – axis above the origin

If C < 0 => Parabola will intersect the y – axis below the origin

If C = 0 => Parabola will pass through the Origin.

>  D (Discriminant):-

The points where the parabola will intersect the X – axis is termed as the X – intercepts of the parabola and at those points the value of function will be zero.

So, If we equate f(x) to 0, we will get a quadratic equation AX2 + B X + C = 0 and the roots or solutions of this quadratic equation are the X – intercepts of the parabola.

So, the solution of the quadratic equation will be given by a formula:

X = (- B ± √D)/2A

Where, D is the Discriminant given as, D = B2 – 4 A C

If D > 0, there will two real and distinct solutions and hence parabola will intersect the X – axis at two different points.

If D = 0, there will be two real and equal solutions and hence parabola will intersect the X – axis at a single point.

And, If D < 0, then there will be no real solutions and hence parabola will not intersect the X – axis.

Now, for the given function f(x) = x2 + b x + 2b2, we will check for the value of A (coefficient of x2), C (Y- intercept) & D(Discriminant).

Value of A (Coefficient of x2) = 1; hence it will be an upward parabola

Value of C (Y-intercept) = 2b2 which will always be positive for any value of b.

Value of D (Discriminant) = b2 – 4 (2b2) = - 7b2 which will always be negative for any value of b; hence it will not intersect the X – axis.

So, with all the conclusions found above, the possible answer is option A

ANS (a)

## Question 11: f(x) = –x4 + 4x2 + 2

g(x) = f(x) + p, where g(x) has 4 distinct roots, which of the following could be the value of p?

a) 3

b) 1

c) 10

d) -3

Given that g(x) = f(x) + p

To find the answer of this question, one must be able to recall the following property:

Adding a positive constant to the function moves up the graph of the function by that amount; whereas subtracting a positive constant move down the graph of the function by that amount.

In this case, if p is positive, graph of g(x) can be obtained by simply moving up the graph of f(x) by p units. If p is negative, graph of g(x) can be obtained by moving down the graph of f(x) by |p| units.

Further, it’s given that y = g(x) has 4 distinct roots.

A root is that value of x for which y becomes 0. So, in this context, y = g(x) having 4 roots means there are 4 different values of x for which g(x) becomes 0, or there are 4 point of intersections of y = g(x) with the X – axis.

In the given situation of the 4 options, y = g(x) can have 4 point of intersection only when p = -3, as that would mean f(x) is moved down by 33 units to give 4 point of intersections with the X – axis.

## Question 12:A magician marks-off on a stick of length 1 yard in thirds and fifths and breaks the stick at the marked points. What is the maximum number of pieces which are equal in length?

a) 2

b) 3

c) 4

d) 5

The figure below shows a 1-yard stick broken into thirds and fifths. Since, we are dealing with fractions, calculating the spacing between the markings and arriving at the above figure can be tedious, because of calculations involved with respect to fractions. To arrive at the answer to this particular question without wasting much time and effort, let’s adopt the following method.

Rather than taking the length of the stick to be 1 yard let’s assume it to be some value which will prevent the calculations from getting into fractions. The easiest value will be the product of the values in the denominators.

Since, we need to break the sticks into thirds (multiples of 1/3) and fifths (multiples of 1/5), let’s assume the length of the stick to be 3 × 5 = 15 yards. Breaking the stick into thirds and fifths, we arrive at this: Thirds are: 1/3 × 15 = 5 & 2/3 × 15 = 10.

Fifths are: 1/5 × 15 = 3, 2/5 × 15 = 6, 3/5 × 15 = 9 & 4/5 × 15 = 12.

So, the stick is broken at the points: 3, 5, 6, 9, 10 & 12.

As it is visible from the figure, number of pieces of:

Length 1 unit = 2 (5 to 6 & 9 to 10)

Length 2 units = 2 (3 to 5 & 10 to 12)

Length 3 units = 3 (0 to 3, 6 to 9 & 12 to 15)

So, the maximum number of pieces that are equal in length are 3, those of length 3 units.

## Question 13: It is given that A, B, C, D and E are integers and A ≤ B ≤ C ≤ D ≤ E. The mean value of A, B, C, D and E is 65 and the median value is 55.What can be the minimum possible value of E?

GIVEN:

• A, B, C, D & E are integers and A ≤ B ≤ C ≤ D ≤ E.
• Mean value of A, B, C, D & E = 65 and the median value is 55.

We have to find the minimum possible value of E.

As C is the middle term so it will be the median.

C = 55

Now, Mean = (A+B+C+D+E)/5 = 65

So, A + B + C + D + E = 325

Putting the value of C,

A + B + 55 + D + E = 325

A + B + D + E = 270 ………………………………………………………………… (1)

Now, whenever the sum of two or more numbers is a constant, then for one of those to be minimum, others have to maximum.

So, For E to be minimum; A, B & D have to be maximum

Now, A & B cannot be more than C, so maximum value of A & B will be equal to C.

A = B = 55

And, as D cannot be more than E so maximum value of D will be equals to E.

D = E

Putting the values of A, B & D in equation (1);

55 + 55 + E + E = 270

2E = 160

E = 80

Hence, Minimum value of E = 80

ANS = 80

## Question 14: Given list of 6 terms: 155, 160, 165, 170, 175 and X. What will be the value of X, so that the standard deviation (S.D.) of the six terms is minimum?

To solve this question, let’s first recall the fact that standard deviation is a measure of dispersion of terms from the mean value of the terms.

Given the list of 6 terms: 155, 160, 165, 170, 175 & X

We can perceive this set as the set consisting of five values: 155, 160, 165, 170, 175; let’s call this set as Set A, and a 6th value, X would be added to the set A.

Indirectly the question becomes: What should be added to Set A such that the standard deviation of the resultant set becomes minimum.

We know that standard deviation of a set that has values far away from the mean is greater than the standard deviation of a set that has values close to the mean.

In this case, to keep the standard minimum, the value to be added (X) should be at the least possible difference from the mean of set A. In other words, X should be at a distance of 0 from the mean, or, X should be equal to the mean of set A.

Hence, X = mean of set A = (155+160+165+170+175)/5 = 165

So, X = 165

## Question 15: Of the 60 cars sold by a dealer in a month, 40 have air-bags, 25 have power-steering, and 12 have both air- bags sand power-steering. How many of the cars in the lot have neither air- bags nor power-steering?

a) 2

b) 7

c) 10

d) 15

Total number of cars sold by a dealer in a month = 60

Number of cars having Air-bags = 40

Number of cars having power steering = 25

Number of cars having both Air bags and power steering = 12

We have to find the number of cars in the lot having neither Air bags nor power steering.

If we represent the data in the table structure:

 Cars With Air Bags Cars without Air bags Total Cars with power Steering 12 13 25 Cars without power steering 28 7 35 Total 40 20 60

So, the number of cars in the lot having neither Air bags nor power steering = 7

ANS: b) = 7

## Question 16: Four drinks – Juice, Coffee, Tea and Soft drink, were rated on a scale of 0 – 5, based on their caffeine and sugar contents, where 0 corresponds to the minimum content and 5 corresponds to the maximum content.

The bar graph above shows caffeine and sugar ratings for the four drinks.

If the given data is represented on a scatter plot, where x – axis would represent sugar content and y – axis would represent caffeine content, then how many points in the scatter plot would lie above the line y = x?

a) 1

b) 2

c) 3

d) 4

If we represent the data on a scatter plot , where x- axis would represent sugar content and y – axis would represent caffeine content , we have to find how many points would lie above the line y =x. From the graph drawn above, the points which will lie on the line y = x will have their y- coordinate equals to the x – coordinate, hence those points for which the caffeine content will be equal to the sugar content.

So, the points which will lie above the line y =x, will have their y- coordinate greater than the x – coordinate, hence those points for which the caffeine content will be more than the sugar content.

So, there is only one drink, coffee for which the caffeine content will be more than the sugar content.

ANS a) = 1

## Question 17:

If a and b are all positive integers and , then what is the value of a2b?

a) 20

b) 50

c) 40

d) 250

We know [(xp yq) r = (xp)r (yq)r]

And [(xp) r = (x)p × r]

Using the above expressions, we can simplify the given expression as below

(a1/4 b1/3)12 = 5000

(a1/4)12 (b1/3)12 = 5000

a3 b4 = 5000.

So, we will expand the term on the right hand side of the equation in the product of its prime factors.

So, we can write 5000 as:

5000 = 23 × 54

So, a3 b4 = 23 × 54

On comparing both the sides,

a3 = 23; b4 = 54

NOTE: If (x)m= (y)m; then

x = y; if m is an odd Integer.

x = (±y); if m is an even integer.

As a3 = 23, then a = 2

And, as b4 = 54, then b = ±5.

But as a and b are given as positive integers,

a = 2; b = 5.

So, a2b = 22 × 5 = 20.

ANS: a) = 20

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