Model Question Paper 3

In the numeral 70542:

Place value of 5 = 500

Face value of 5 = 5

Difference between place value and face value = 500 - 5 = 495

Hence, option 3 is the correct option.

Successor of 99 = 99 + 1 = 100

Predecessor of 101 = 101 - 1 = 100

Sum = 100 + 100 = 200

Hence, option 2 is the correct option.

If two numbers are co-prime, then their HCF = 1.

We know that, product of two numbers = HCF × LCM

⇒ $x ×$y = 1 × LCM

⇒ LCM = $xy$

Hence, option 1 is the correct option.

Let us check each option:

1. |15 - 6| = |9| = 9, but |15| + |-6| = 15 + 6 = 21. So 9 ≠ 21. False.

2. Additive inverse of -3 is 3, since -3 + 3 = 0. True.

3. -1 lies on the left of 0 on the number line, not on the right. False.

4. -4 is less than -3 (i.e., -4 < -3). False.

Hence, option 2 is the correct option.

$\frac{360}{540} = \frac{360 \div 180}{540 \div 180} = \frac{2}{3}$

Now let us check each option:

1. $\frac{4}{6} = \frac{4 \div 2}{6 \div 2} = \frac{2}{3}$ ✓

2. $\frac{3}{2}$ — not equal to $\frac{2}{3}$

3. $\frac{3}{4}$ — not equal to $\frac{2}{3}$

4. $\frac{9}{18} = \frac{1}{2}$ — not equal to $\frac{2}{3}$

Hence, option 1 is the correct option.

A number is divisible by 6 if it is divisible by both 2 and 3.

1. 5372: Ends in 2, so divisible by 2. Sum of digits = 5 + 3 + 7 + 2 = 17, not divisible by 3. So not divisible by 6.

2. 6495: Ends in 5, so not divisible by 2. So not divisible by 6.

3. 7632: Ends in 2, so divisible by 2. Sum of digits = 7 + 6 + 3 + 2 = 18, divisible by 3. So divisible by 6. ✓

4. 7568: Ends in 8, so divisible by 2. Sum of digits = 7 + 5 + 6 + 8 = 26, not divisible by 3. So not divisible by 6.

Hence, option 3 is the correct option.

Statement I: Factors of 12 = 1, 2, 3, 4, 6, 12. Factors of 18 = 1, 2, 3, 6, 9, 18.

Common factors = 1, 2, 3, 6. Highest common factor = 6.

So HCF of 12 and 18 is 6. Statement I is true.

Statement II: From the list of common factors above, 6 is a common factor of 12 and 18. Statement II is true.

Hence, option 3 is the correct option.

Statement I: 1 m = 100 cm

Ratio of 5 cm to 1 m = $\frac{5}{100} = \frac{1}{20}$ or 1 : 20, not 5 : 1. Statement I is false.

Statement II: Sum of terms of the ratio = 5 + 1 = 6

First friend gets = $\frac{5}{6} \times 30 = 25$ pencils

Second friend gets = $\frac{1}{6} \times 30 = 5$ pencils

So they get 25 and 5 pencils respectively. Statement II is true.

Hence, option 2 is the correct option.

Greatest 4-digit number: To get the greatest number, place the largest available digits at higher places. The digit 4 must be at tens place.

Thousands place: largest digit = 9

Hundreds place: next largest digit (different from 9 and 4) = 8

Tens place: 4 (given)

Ones place: next largest digit (different from 9, 8, 4) = 7

So, the greatest 4-digit number = 9847

Smallest 4-digit number: To get the smallest number, place the smallest available digits at higher places. The digit at thousands place cannot be 0.

Thousands place: smallest non-zero digit = 1

Hundreds place: smallest digit (different from 1 and 4) = 0

Tens place: 4 (given)

Ones place: next smallest digit (different from 1, 0, 4) = 2

So, the smallest 4-digit number = 1042

We use repeated division by prime numbers:

$$\begin{array}{l|r} 2 & 980 \\ \hline 2 & 490 \\ \hline 5 & 245 \\ \hline 7 & 49 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$

Hence, prime factorisation of 980 = 2 × 2 × 5 × 7 × 7 = 2² × 5 × 7²

Solving,

-15 - (-2) - 71 - (-8) + 6

= -15 + 2 - 71 + 8 + 6 [since -(-2) = +2 and -(-8) = +8]

= (-15 - 71) + (2 + 8 + 6)

= -86 + 16

= -70.

Hence, -15 - (-2) - 71 - (-8) + 6 = -70.

From the figure, the rectangle is divided into 8 equal parts.

Out of that one full and three half rectangle are shaded.

Number of shaded parts = $\frac{1}{2} + \frac{1}{2} + \frac{1}{2} + 1 = \frac{3}{2} + 1 = \frac{5}{2}$

∴ Fraction = $\frac{\text{Number of shaded parts}}{\text{Total number of parts}} = \frac{\frac{5}{2}}{8}$

Hence, required fraction = $\frac{5}{16}$.

$7\frac{3}{40} = 7 + \frac{3}{40}$

Multiplying the numerator and denominator of $\frac{3}{40}$ by 25 to make the denominator 1000:

$\frac{3}{40} = \frac{3 \times 25}{40 \times 25} = \frac{75}{1000} = 0.075$

∴ $7\frac{3}{40} = 7 + 0.075 = 7.075$

Length of pencil = 14 cm = 14 × 10 mm = 140 mm

Diameter of pencil = 7 mm

(Converting both quantities to the same unit, i.e., mm.)

Ratio of diameter to length = $\frac{7}{140} = \frac{1}{20}$ = 1 : 20.

(i) Rounding off to nearest ten:

2459 → 2460 (since 9 is rounded up)

653 → 650 (since 3 is rounded down)

Estimated product = 2460 × 650 = 15,99,000

(ii) Rounding off to nearest hundred:

2459 → 2500 (since the tens digit is 5)

653 → 700 (since the tens digit is 5)

Estimated product = 2500 × 700 = 17,50,000

Using distributive law of multiplication over addition:

257 × 1007 = 257 × (1000 + 7)

= 257 × 1000 + 257 × 7

= 2,57,000 + 1,799

= 2,58,799

Hence, 257 × 1007 = 2,58,799.

Subtract (-3) from (-8), i.e. (-8) - (-3)

To subtract a negative number,

We start at the first number and move to the right (towards larger values) by the value of the second number.

On the number line, we start at -8 and move 3 units to the right (since we are adding 3):

Starting at -8, moving 3 units to the right we reach -5.

∴ (-8) - (-3) = -5.

The greatest number that will divide 76, 113 and 186 leaving remainders 4, 5 and 6 respectively is the HCF of (76 - 4), (113 - 5) and (186 - 6), i.e., HCF of 72, 108 and 180.

Prime factorisation of 72 :

$$\begin{array}{l|r} 2 & 72 \\ \hline 2 & 36 \\ \hline 2 & 18 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

Prime factorisation of 108 :

$$\begin{array}{l|r} 2 & 108 \\ \hline 2 & 54 \\ \hline 3 & 27 \\ \hline 3 & 9 \\ \hline 3 & 3 \\ \hline & 1 \end{array}$$

Prime factorisation of 180 :

$$\begin{array}{l|r} 2 & 180 \\ \hline 2 & 90 \\ \hline 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

72 = 2 × 2 × 2 × 3 × 3 = 2³ × 3²

108 = 2 × 2 × 3 × 3 × 3 = 2² × 3³

180 = 2 × 2 × 3 × 3 × 5 = 2² × 3² × 5

Since HCF is found by taking only the common prime factors with the smallest powers, we take 2² and 3², which are common to 72, 108 and 180.

HCF = 2² × 3² = 4 × 9 = 36

Hence, the required greatest number is 36..

To arrange in descending order, we go from the greatest to the smallest.

For positive integers: 2376 > 2367 > 245

For negative integers, the one with the smaller absolute value is greater:

-311 > -506 > -509 (since 311 < 506 < 509)

Combining (all positives are greater than all negatives):

2376, 2367, 245, -311, -506, -509

To compare the fractions, we convert them to like fractions by taking the LCM of denominators.

LCM of 12, 4, 8, 6:

12 = 2² × 3

4 = 2²

8 = 2³

6 = 2 × 3

LCM = 2³ × 3 = 24

Converting to like fractions with denominator 24:

$$\begin{aligned}\frac{5}{12} = \frac{5 \times 2}{12 \times 2} = \frac{10}{24} \\[1em] \frac{1}{4} = \frac{1 \times 6}{4 \times 6} = \frac{6}{24} \\[1em] \frac{7}{8} = \frac{7 \times 3}{8 \times 3} = \frac{21}{24} \\[1em] \frac{5}{6} = \frac{5 \times 4}{6 \times 4} = \frac{20}{24} \\[1em]\end{aligned}$$

Now,

$\frac{6}{24} \lt \frac{10}{24} \lt \frac{20}{24} \lt \frac{21}{24}$

∴ Ascending order: $\frac{1}{4}, \frac{5}{12}, \frac{5}{6}, \frac{7}{8}$

Converting mixed fractions to improper fractions:

$3\frac{5}{6} = \frac{3 \times 6 + 5}{6} = \frac{23}{6}$

$4\frac{3}{4} = \frac{4 \times 4 + 3}{4} = \frac{19}{4}$

$1\frac{3}{8} = \frac{1 \times 8 + 3}{8} = \frac{11}{8}$

So, the expression becomes:

$\frac{23}{6} + \frac{19}{4} - 5 - \frac{11}{8}$

LCM of 6, 4, 1, 8 = 24

Converting to like fractions with denominator 24:

$$\begin{aligned}\frac{23}{6} = \frac{23 \times 4}{24} = \frac{92}{24}\\[1em] \frac{19}{4} = \frac{19 \times 6}{24} = \frac{114}{24}\\[1em] 5 = \frac{5 \times 24}{24} = \frac{120}{24}\\[1em] \frac{11}{8} = \frac{11 \times 3}{24} = \frac{33}{24}\\[1em] \text {Sum} = \frac{92}{24} + \frac{114}{24} - \frac{120}{24} - \frac{33}{24}\\[1em] = \frac{92 + 114 - 120 - 33}{24}\\[1em] = \frac{206 - 153}{24}\\[1em] = \frac{53}{24}\\[1em] = 2\frac{5}{24}.\end{aligned}$$

Hence, $3\frac{5}{6} + 4\frac{3}{4} - 5 - 1\frac{3}{8} = 2\frac{5}{24}$

Using the digits 3, 0, 8 without repetition:

1-digit natural numbers: 3, 8 (0 is not a natural number)

2-digit natural numbers (the first digit cannot be 0):

Starting with 3: 30, 38

Starting with 8: 80, 83

So, 2-digit numbers: 30, 38, 80, 83

3-digit natural numbers (the first digit cannot be 0):

Starting with 3: 308, 380

Starting with 8: 803, 830

So, 3-digit numbers: 308, 380, 803, 830

All possible natural numbers: 3, 8, 30, 38, 80, 83, 308, 380, 803, 830

Sum of all these numbers:

= 3 + 8 + 30 + 38 + 80 + 83 + 308 + 380 + 803 + 830

= 11 + 68 + 163 + 688 + 1633

Sum = 2563

Ages of Munna and Munni = 14 years and 10 years

Ratio of their ages = 14 : 10 = $\frac{14}{10} = \frac{7}{5}$ or 7 : 5

Sum of the terms of the ratio = 7 + 5 = 12

Total money to be divided = ₹ 180

Munna gets $(= \frac{7}{12} \times ₹180 = ₹105)$

Munni gets $(= \frac{5}{12} \times ₹180 = ₹75)$

Hence, Munna gets = ₹ 105 and Munni gets = ₹ 75.

First ratio: 15 cm to 3 m

Converting to the same unit: 3 m = 3 × 100 cm = 300 cm

Ratio of 15 cm to 3 m = $\frac{15}{300} = \frac{1}{20}$ or 1 : 20

Second ratio: 15 seconds to 3 minutes

Converting to the same unit: 3 minutes = 3 × 60 seconds = 180 seconds

Ratio of 15 seconds to 3 minutes = $\frac{15}{180} = \frac{1}{12}$ or 1 : 12

Since $\frac{1}{20} \neq \frac{1}{12}$, the two ratios are not equal.

No, the given ratios do not form a proportion.

Cloth needed to stitch one shirt = 2 m 25 cm = (2 × 100 + 25) cm = 225 cm

Total cloth available = 30 m = 30 × 100 cm = 3000 cm

Number of shirts that can be stitched

= $\frac{\text{Total cloth}}{\text{Cloth needed per shirt}} = \frac{3000}{225}$

On dividing 3000 by 225:

$$\begin{array}{l}225\overline{\smash{\big)}3000\smash{\big(}} 13 \\ \phantom{ac)}\phantom{}\underline{-225} \\ \phantom{{x^2 =))}} 750 \\ \phantom{ac+}\phantom{}\underline{-675} \\ \phantom{{x^2 =)+)}} 75 \end{array}$$

3000 ÷ 225 → quotient = 13, remainder = 3000 - (225 × 13) = 3000 - 2925 = 75

So, 13 shirts can be stitched and 75 cm of cloth will remain.

Hence, 13 shirts can be stitched and 75 cm of cloth will remain.

The bells will ring together again after a time which is the LCM of 40, 30, 36 and 45 minutes.

LCM of 40, 30, 36 and 45:

$$\begin{array}{l|rrrr} 2 & 40, & 30, & 36, & 45 \\ \hline 2 & 20, & 15, & 18, & 45 \\ \hline 2 & 10, & 15, & 9, & 45 \\ \hline 3 & 5, & 15, & 9, & 45 \\ \hline 3 & 5, & 5, & 3, & 15 \\ \hline 5 & 5, & 5, & 1, & 5 \\ \hline & 1, & 1, & 1, & 1 \end{array}$$

LCM = 2³ × 3² × 5 = 8 × 9 × 5 = 360

So, the bells will ring together again after 360 minutes.

360 minutes = $\frac{360}{60}$ hours = 6 hours

The bells start ringing at 9 A.M.

They will ring together again at 9 A.M. + 6 hours = 3 P.M.

Hence, the bells will ring together simultaneously at 3 P.M.

Cost of fruits = ₹ $27\frac{3}{4} = ₹ \frac{111}{4}$

Cost of vegetables = ₹ $10\frac{1}{2} = ₹\frac{21}{2}$

Total cost = ₹ $\left(\frac{111}{4} + \frac{21}{2}\right)$

LCM of 4 and 2 = 4

= ₹ $\left(\frac{111}{4} + \frac{42}{4}\right)$

= ₹ $\frac{153}{4}$

= ₹ $38\frac{1}{4}$

Money given to the shopkeeper = ₹ 50

Money to get back = ₹ 50 - ₹ $\frac{153}{4}$

= ₹ $\left(\frac{200}{4} - \frac{153}{4}\right)$

= ₹ $\frac{47}{4}$

Hence, the money i will get back = ₹ $11\frac{3}{4}$.

Converting all weights into grams (1 kg = 1000 g):

Total weight of vegetables = 10 kg = 10000 g

Weight of potatoes = 3 kg 450 g = 3000 + 450 = 3450 g

Weight of onions = 2 kg 10 g = 2000 + 10 = 2010 g

Weight of tomatoes = 1 kg 750 g = 1000 + 750 = 1750 g

Total weight of potatoes, onions and tomatoes = 3450 + 2010 + 1750

= 7210 g

= 7 kg 210 g

∴ Weight of green vegetables = Total weight - (weight of potatoes + onions + tomatoes)

= 10000 g - 7210 g

= 2790 g = 2000 g + 790 g

Hence, weight of green vegetables = 2 kg 790 g.

The finish line is at the 100 m mark. A position of +10 means the runner is 10 m in front of the starting line, so the runner is closer to the finish line. A position of -5 means the runner is 5 m behind the starting line, so the runner is farther from the finish line.

Distance to the finish line for each runner = 100 - (starting position)

(i) The runner with the lowest, that is, the most negative starting position, is farthest from the finish line.

Arranging the starting positions:

-20 < -3 < 0 < +5 < +10 < +15

The lowest starting position is -20, which belongs to Rishi.

Hence, Rishi starts farthest from the finish line.

(ii) Tarush's starting position = +15

Distance Tarush has to run to reach the finish line = 100 - 15 = 85 m

Hence, Tarush has to run 85 m to reach the finish line.

(iii) Mohan's starting position = -3

Distance Mohan has to run to reach the finish line = 100 - (-3) = 100 + 3 = 103 m

Hence, Mohan has to run 103 m to reach the finish line.

(iv) Rishi's starting position = -20

Peter's starting position = +10

Distance between their starting positions = 10 - (-20) = 10 + 20 = 30 m

Hence, the distance between the starting positions of Rishi and Peter is 30 m.