Hard GCSE Dividing in a ratio Questions

Challenging, exam-style GCSE Dividing in a ratio questions with worked solutions. Stretch yourself on the hardest difference between shares, finding the whole, working backwards, three-part ratio problems.

difference between sharesfinding the wholeworking backwardsthree-part ratiosharing in a ratiotwo-stage sharing
GCSE Foundation34 questionsStep-by-step solutions
Question 1
6 markschallenging
Two numbers are in the ratio 4:74 : 7. When 55 is added to the smaller number and 55 is subtracted from the larger number, the two results are in the ratio 3:43 : 4. Work out the two original numbers.
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Worked solution

  1. Set up the unknown

    k=value of one partk = \text{value of one part}

    The ratio fixes the shape of the numbers but not their size, so introduce one unknown.

  2. Write the two numbers

    smaller=4k,larger=7k\text{smaller} = 4k, \quad \text{larger} = 7k

    Every pair in the ratio 4:74 : 7 has this form.

  3. Apply the change to the smaller number

    4k+54k + 5

    Five is ADDED to the smaller number.

  4. Apply the change to the larger number

    7k−57k - 5

    Five is SUBTRACTED from the larger number — different operations, so read carefully.

  5. Write the new ratio

    (4k+5):(7k−5)=3:4(4k + 5) : (7k - 5) = 3 : 4

    The two results are in the ratio 3:43 : 4.

  6. Turn it into a fraction equation

    4k+57k−5=34\frac{4k + 5}{7k - 5} = \frac{3}{4}

    Equal ratios mean equal fractions.

  7. Cross-multiply

    4(4k+5)=3(7k−5)4(4k + 5) = 3(7k - 5)

    Multiply both sides by 4(7k−5)4(7k - 5).

  8. Expand the left-hand side

    16k+2016k + 20

    Multiply each term inside the bracket by 44.

  9. Expand the right-hand side

    21k−1521k - 15

    Multiply each term by 33. Note that 3×(−5)=−153 \times (-5) = -15.

  10. Write the equation out

    16k+20=21k−1516k + 20 = 21k - 15

    Both sides are now expanded.

  11. Collect the k terms

    20+15=21k−16k20 + 15 = 21k - 16k

    Move the kk terms one way and the numbers the other.

  12. Solve for one part

    35=5k⇒k=735 = 5k \Rightarrow k = 7

    One part is 77.

  13. Work out the two numbers

    4×7=28,7×7=494 \times 7 = 28, \quad 7 \times 7 = 49

    The original numbers are 2828 and 4949.

  14. Check the original ratio

    28:49=4:728 : 49 = 4 : 7

    Dividing both by 77 gives 4:74 : 7. ✓\checkmark

  15. Check the new ratio

    (28+5):(49−5)=33:44=3:4(28 + 5) : (49 - 5) = 33 : 44 = 3 : 4

    Dividing 33:4433 : 44 by 1111 gives 3:43 : 4 — exactly as required. ✓\checkmark

Answer
28 and 4928 \text{ and } 49
Question 2
5 markschallenging
A club has men and women members in the ratio 5:45 : 4. There are 180180 members. Then 20%20\% of the men leave the club and 1010 more women join. Write the new ratio of men to women in its simplest form.
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Worked solution

  1. Add the parts of the ratio

    5+4=95 + 4 = 9

    The 180180 members are split into 99 equal parts.

  2. Check the division works exactly

    180÷9=20180 \div 9 = 20

    One part is 2020 members.

  3. Work out the number of men

    5×20=1005 \times 20 = 100

    There are 100100 men.

  4. Work out the number of women

    4×20=804 \times 20 = 80

    There are 8080 women.

  5. Check the members add back to the total

    100+80=180100 + 80 = 180

    The men and women total 180180 members. ✓\checkmark

  6. Work out how many men leave

    20% of 100=20100×100=2020\% \text{ of } 100 = \frac{20}{100} \times 100 = 20

    Twenty per cent of the MEN — not of the whole club — leave.

  7. Work out the men remaining

    100−20=80100 - 20 = 80

    There are 8080 men left.

  8. Work out the new number of women

    80+10=9080 + 10 = 90

    Ten women join, giving 9090 women.

  9. Write the new ratio

    80:9080 : 90

    Men to women, in that order.

  10. Find the highest common factor

    HCF(80,90)=10\text{HCF}(80, 90) = 10

    Both numbers divide by 1010.

  11. Simplify the ratio

    80÷10:90÷10=8:980 \div 10 : 90 \div 10 = 8 : 9

    The new ratio is 8:98 : 9.

  12. Check the simplified ratio

    HCF(8,9)=1\text{HCF}(8, 9) = 1

    88 and 99 share no common factor, so 8:98 : 9 is fully simplified. ✓\checkmark

  13. Check the new membership total

    80+90=17080 + 90 = 170

    The club now has 170170 members: 180−20+10=170180 - 20 + 10 = 170. ✓\checkmark

  14. Notice the ratio has flipped over

    5:4→8:95 : 4 \rightarrow 8 : 9

    The men used to outnumber the women; now the women outnumber the men.

  15. Watch the common error

    20% of 180=3620\% \text{ of } 180 = 36

    Taking 20%20\% of ALL the members gives 3636, which is not what the question says — the percentage applies only to the men.

Answer
8:98 : 9
Question 3
6 markschallenging
Dan, Eve and Fay share a sum of money in the ratio 5:3:25 : 3 : 2. Dan then gives 15\frac{1}{5} of his share to Fay. Fay now has £84. Work out the total amount of money that was shared.
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Worked solution

  1. Let one part be k pounds

    Dan=5k,Eve=3k,Fay=2k\text{Dan} = 5k, \quad \text{Eve} = 3k, \quad \text{Fay} = 2k

    Writing every share in terms of one part is the key move.

  2. Work out what Dan gives away

    15×5k=k\frac{1}{5} \times 5k = k

    A fifth of Dan’s 5k5k is exactly kk — one whole part. This is why the ratio 5:3:25 : 3 : 2 was chosen.

  3. Write what Dan has left

    5k−k=4k5k - k = 4k

    Dan keeps 4k4k pounds.

  4. Write what Fay has now

    2k+k=3k2k + k = 3k

    Fay’s 2k2k grows by kk to become 3k3k pounds.

  5. Set Fay’s new amount equal to £84

    3k=843k = 84

    The question says Fay now has £84.

  6. Solve for one part

    k=84÷3=28k = 84 \div 3 = 28

    One part is worth £28\pounds 28.

  7. Count the parts in the whole

    5+3+2=10 parts5 + 3 + 2 = 10 \text{ parts}

    The original total was 1010 parts.

  8. Work out the total

    10×28=28010 \times 28 = 280

    The total shared was £280\pounds 280.

  9. Work out Dan’s original share

    5×28=1405 \times 28 = 140

    Dan started with £140\pounds 140.

  10. Work out Eve’s share

    3×28=843 \times 28 = 84

    Eve receives £84\pounds 84 and never changes.

  11. Work out Fay’s original share

    2×28=562 \times 28 = 56

    Fay started with £56\pounds 56.

  12. Check the original shares add to the total

    140+84+56=280140 + 84 + 56 = 280

    The three original shares total £280. ✓\checkmark

  13. Check the transfer

    15×140=28,56+28=84\frac{1}{5} \times 140 = 28, \quad 56 + 28 = 84

    Dan gives away £28 and Fay ends with £84 — exactly as the question states. ✓\checkmark

  14. Check the money after the transfer

    112+84+84=280112 + 84 + 84 = 280

    Dan has £112 left, Eve £84 and Fay £84 — still £280 in total. ✓\checkmark

  15. Watch the common error

    84÷2=4284 \div 2 = 42

    Treating Fay’s £84 as her ORIGINAL 22-part share gives k=42k = 42 and a total of £420, which fails the check.

Answer
£280\pounds 280
Question 4
6 markschallenging
A 2.42.4 kg bag of flour is divided between three bakers in the ratio 3:4:53 : 4 : 5. The baker with the smallest portion then uses 250250 g of her flour. Write the new ratio of the three bakers’ remaining flour in its simplest form.
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Worked solution

  1. Spot the units problem

    2.4 kg vs 250 g2.4 \text{ kg} \text{ vs } 250 \text{ g}

    The bag is in kilograms but the flour used is in grams — convert everything to grams first.

  2. Convert the bag to grams

    2.4×1000=2400 g2.4 \times 1000 = 2400 \text{ g}

    The bag holds 24002400 g of flour.

  3. Add the parts of the ratio

    3+4+5=123 + 4 + 5 = 12

    The flour is split into 1212 equal parts.

  4. Find one part

    2400÷12=2002400 \div 12 = 200

    One part is 200200 g.

  5. Work out the smallest portion

    3×200=6003 \times 200 = 600

    The smallest portion is 600600 g.

  6. Work out the middle portion

    4×200=8004 \times 200 = 800

    The middle portion is 800800 g.

  7. Work out the largest portion

    5×200=10005 \times 200 = 1000

    The largest portion is 10001000 g =1= 1 kg.

  8. Check the portions add back to the bag

    600+800+1000=2400600 + 800 + 1000 = 2400

    The three portions total 24002400 g. ✓\checkmark

  9. Apply the change

    600−250=350600 - 250 = 350

    The smallest portion drops to 350350 g. The other two are untouched.

  10. Write the new amounts as a ratio

    350:800:1000350 : 800 : 1000

    Keep the bakers in the same order as the original ratio.

  11. Find the highest common factor

    HCF(350,800,1000)=50\text{HCF}(350, 800, 1000) = 50

    350=50×7350 = 50 \times 7, 800=50×16800 = 50 \times 16 and 1000=50×201000 = 50 \times 20.

  12. Simplify the ratio

    350÷50:800÷50:1000÷50=7:16:20350 \div 50 : 800 \div 50 : 1000 \div 50 = 7 : 16 : 20

    The new ratio is 7:16:207 : 16 : 20.

  13. Check the simplified ratio cannot be reduced further

    HCF(7,16,20)=1\text{HCF}(7, 16, 20) = 1

    77 is prime and does not divide 1616 or 2020, so the ratio is in its simplest form. ✓\checkmark

  14. Check the new total

    350+800+1000=2150,2400−250=2150350 + 800 + 1000 = 2150, \quad 2400 - 250 = 2150

    The remaining flour is 21502150 g by both routes. ✓\checkmark

  15. Note why the ratio changed

    3:4:5→7:16:203 : 4 : 5 \rightarrow 7 : 16 : 20

    Removing flour from only ONE portion changes the ratio; the parts no longer share a common multiplier of 200200.

Answer
7:16:207 : 16 : 20
Question 5
6 markschallenging
Two children share some sweets in the ratio 7:37 : 3. The first child then gives 88 sweets to the second child, and now they have sweets in the ratio 3:23 : 2. How many sweets were shared altogether?
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Worked solution

  1. Plan the method

    one part=k→equation→total\text{one part} = k \rightarrow \text{equation} \rightarrow \text{total}

    Giving sweets away changes the ratio, so the size must be found with algebra.

  2. Write the original amounts

    first=7k,second=3k\text{first} = 7k, \quad \text{second} = 3k

    Let one part be kk sweets.

  3. Apply the transfer to the first child

    7k−87k - 8

    The first child gives 88 sweets away.

  4. Apply the transfer to the second child

    3k+83k + 8

    The second child RECEIVES those same 88 sweets — note the plus sign.

  5. Write the new ratio

    (7k−8):(3k+8)=3:2(7k - 8) : (3k + 8) = 3 : 2

    After the transfer the sweets are in the ratio 3:23 : 2.

  6. Turn it into a fraction equation

    7k−83k+8=32\frac{7k - 8}{3k + 8} = \frac{3}{2}

    Equal ratios mean equal fractions.

  7. Cross-multiply

    2(7k−8)=3(3k+8)2(7k - 8) = 3(3k + 8)

    Multiply out to clear the fractions.

  8. Expand both brackets

    14k−16=9k+2414k - 16 = 9k + 24

    Take care with the signs.

  9. Collect the k terms

    14k−9k=24+1614k - 9k = 24 + 16

    Move 9k9k left and −16-16 right.

  10. Solve for one part

    5k=40⇒k=85k = 40 \Rightarrow k = 8

    One part is 88 sweets.

  11. Work out the original amounts

    7×8=56,3×8=247 \times 8 = 56, \quad 3 \times 8 = 24

    The children started with 5656 and 2424 sweets.

  12. Work out the total

    56+24=8056 + 24 = 80

    There were 8080 sweets altogether.

  13. Check the amounts after the transfer

    56−8=48,24+8=3256 - 8 = 48, \quad 24 + 8 = 32

    They now have 4848 and 3232 sweets.

  14. Check the new ratio

    48:32=3:248 : 32 = 3 : 2

    Dividing both by 1616 gives 3:23 : 2 — exactly as the question says. ✓\checkmark

  15. Check the total is unchanged

    48+32=8048 + 32 = 80

    The transfer moves sweets around but does not change the total, which is still 8080. ✓\checkmark

Answer
80 sweets80 \text{ sweets}

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